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\title{Further trigonometry }

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\date{}


\begin{document}
\maketitle
Trigonometry begins with the study of relationships between sides and angles in a right-angled triangle.

In this chapter, we will review the basics of the trigonometry of right-angled triangles, look at applications to three-dimensional problems, and extend our study of trigonometry to triangles that are not right-angled.
\includegraphics[max width=\textwidth, center]{2024_02_22_e203939eb913b173edc2g-01}

\section*{12 Review of the basic trigonometric ratios}
By similarity, the ratio of any two sides in a right-angled triangle is always the same, once we have fixed the angles.

We choose one of the two acute angles and call it the reference angle.

The side opposite the reference angle is called the opposite, the

\includegraphics[max width=\textwidth, center]{2024_02_22_e203939eb913b173edc2g-02(1)}
side opposite the right angle is called the hypotenuse and the remaining side, which is between the reference angle and the right-angle, is called the adjacent.

The three basic trigonometric ratios are the sine, cosine and tangent ratios.

\[
\sin \theta=\frac{\text { opposite }}{\text { hypotenuse }} \quad \cos \theta=\frac{\text { adjacent }}{\text { hypotenuse }} \quad \tan \theta=\frac{\text { opposite }}{\text { adjacent }}
\]

You should learn the three ratios for sine, cosine and tangent by heart and remember them. A simple mnemonic is:

\section*{SOHCAHTOA}
for Sine: Opposite/Hypotenuse, Cosine: Adjacent/Hypotenuse, Tangent: Opposite/Adjacent

\section*{Complementary angles}
In the diagram, the angles at \(A\) and \(B\) are complementary; that is, they add to \(90^{\circ}\).

The side opposite \(A\) is the side adjacent to \(B\) and vice versa. Hence, the sine of \(\theta\) is the cosine of \(\left(90^{\circ}-\theta\right)\) and vice versa.

\[
\begin{aligned}
& \sin \theta=\cos \left(90^{\circ}-\theta\right) \\
& \cos \theta=\sin \left(90^{\circ}-\theta\right)
\end{aligned}
\]

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-02(2)}
\end{center}

For example, \(\sin 60^{\circ}=\cos 30^{\circ}\) and \(\cos 10^{\circ}=\sin 80^{\circ}\).

\section*{Example 1}
Write down the sine, cosine and tangent ratios for the angle \(\theta\) in this triangle.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-02}
\end{center}

\section*{Solution}
\[
\sin \theta=\frac{12}{13} \quad \cos \theta=\frac{5}{13} \quad \tan \theta=\frac{12}{5}
\]

Once a reference angle is given, the approximate numerical value of each of the three ratios can be obtained from a calculator. We can use this idea to find unknown sides in a right-angled triangle.

\section*{Example 2}
Find, correct to two decimal places, the value of the pronumeral in each triangle.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-03(2)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-03(1)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-03(3)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-03}
\end{center}

\section*{Solution}
a \(\sin 15^{\circ}=\frac{\text { opposite }}{\text { hypotenuse }}\)

\(\sin 15^{\circ}=\frac{x}{8}\)

\(x=8 \times \sin 15^{\circ}\)

\(\approx 2.07\)

c \(\tan 43^{\circ}=\frac{\text { opposite }}{\text { adjacent }}\)

\(\tan 43^{\circ}=\frac{d}{6.2}\)

\(d=6.2 \tan 43^{\circ}\)

\(\approx 5.78\) b \(\cos 28^{\circ}=\frac{\text { adjacent }}{\text { hypotenuse }}\)

\(\cos 28^{\circ}=\frac{a}{12.2}\)

\(a=12.2 \times \cos 28^{\circ}\)

\(\approx 10.77\)

d

\(\tan 37^{\circ}=\frac{\text { opposite }}{\text { adjacent }}\)

\(\tan 37^{\circ}=\frac{6}{x}\)

\(x \tan 37^{\circ}=6\)

\(x=\frac{6}{\tan 37^{\circ}}\)

\(\approx 7.96\)

\section*{Finding angles}
In order to apply trigonometry to finding angles rather than side lengths in right-angled triangles, we need to be able to go from the value of sine, cosine or tangent back to the angle.

What is the acute angle whose sine is 0.5 ?

The calculator gives \(\sin 30^{\circ}=0.5\), so we write \(\sin ^{-1} 0.5=30^{\circ}\).

The opposite process of finding the sine of an angle is to find the inverse sine of a number.

When \(\theta\) is an acute angle, the statement \(\sin ^{-1} x=\theta\) means \(\sin \theta=x\).

This notation is standard, but is rather misleading. The index -1 does NOT mean one over, as it normally does in algebra. To help you avoid confusion, you should always read \(\sin ^{-1} x\) as inverse sine of \(x\) and \(\tan ^{-1} x\) as inverse tan of \(x\), and so on.

For example, the calculator gives \(\cos ^{-1} 0.8192 \approx 35^{\circ}(\mathrm{read}\) this as inverse cosine of 0.8192 is approximately \(35^{\circ}\) ).

\section*{Example 3}
Calculate the value of \(\theta\), correct to one decimal place.

\(\mathbf{a}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-04(1)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-04(10)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-04(2)}
\end{center}

\section*{Solution}
a \(\sin \theta=\frac{6}{11}\)
b \(\cos \theta=\frac{8.2}{14}\)
c \(\tan \theta=\frac{7}{12}\)
\(\theta=\sin ^{-1}\left(\frac{6}{11}\right)\)
\(\theta=\cos ^{-1}\left(\frac{8.2}{14}\right)\)
\(\theta=\tan ^{-1}\left(\frac{7}{12}\right)\)
\(\approx 33.1^{\circ}\)
\(\approx 54.1^{\circ}\)
\(\approx 30.3^{\circ}\)

\section*{Exercise 12A}
1 Calculate the value of each pronumeral, correct to two decimal places.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-04(8)}
\end{center}

d

\begin{center}
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\end{center}

b

\begin{center}
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\end{center}

e

\begin{center}
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\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-04(7)}
\end{center}

f

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-04}
\end{center}

2 Calculate the value of the pronumeral, correct to two decimal places.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-04(5)}
\end{center}

\(\mathbf{c}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-04(3)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-04(4)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-04(11)}
\end{center}

Example 3 Calculate the value of \(\theta\), correct to one decimal place.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-05(7)}
\end{center}

c

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\end{center}

e

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\end{center}

b

\begin{center}
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\end{center}

d

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\end{center}

f

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-05(3)}
\end{center}

4 Calculate the value of each pronumeral. Give side lengths correct to two decimal places and angles correct to one decimal place.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-05(12)}
\end{center}

d

\begin{center}
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\end{center}

g

\begin{center}
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\end{center}

b

\begin{center}
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\end{center}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-05(6)}
\end{center}

h

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-05(13)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-05(11)}
\end{center}

f

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-05(8)}
\end{center}

i

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-05(10)}
\end{center}

5 Find all sides, correct to two decimal places, and all angles, correct to one decimal place.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-06(3)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-06}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-06(1)}
\end{center}

\section*{12 Exact values}
The trigonometric ratios for the angles \(30^{\circ}, 45^{\circ}\) and \(60^{\circ}\) occur very frequently and can be expressed using surds.

The value of the trigonometric ratios for \(45^{\circ}\) can be found from the diagram opposite. It is an isosceles triangle with shorter sides 1.

The values of the trigonometric ratios for \(30^{\circ}\) and \(60^{\circ}\) can be found by drawing an altitude in an equilateral triangle.

The values are given in the table.

\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
\(\theta\) & \(\sin \theta\) & \(\cos \theta\) & \(\tan \theta\) \\
\hline
\(30^{\circ}\) & \(\frac{1}{2}\) & \(\frac{\sqrt{3}}{2}\) & \(\frac{1}{\sqrt{3}}\) \\
\hline
\(45^{\circ}\) & \(\frac{1}{\sqrt{2}}\) & \(\frac{1}{\sqrt{2}}\) & 1 \\
\hline
\(60^{\circ}\) & \(\frac{\sqrt{3}}{2}\) & \(\frac{1}{2}\) & \(\sqrt{3}\) \\
\hline
\end{tabular}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-06(5)}
\end{center}

1

Check the details in the triangles and the entries in the table. You can either learn the table or remember the diagrams to construct the table.

\section*{Example 4}
Find the exact value of \(x\).

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-06(4)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-06(2)}
\end{center}

\section*{Solution}
a We have \(\cos 30^{\circ}=\frac{x}{8}\)

\[
\begin{aligned}
x & =8 \cos 30^{\circ} \\
& =8 \times \frac{\sqrt{3}}{2} \\
& =4 \sqrt{3}
\end{aligned}
\]

b We have \(\tan 60^{\circ}=\frac{6}{x}\)

so

\[
\frac{6}{x}=\sqrt{3}
\]

Hence

\[
\begin{aligned}
x & =\frac{6}{\sqrt{3}} \\
& =\frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \text { (Rationalise the denominator.) } \\
& =2 \sqrt{3}
\end{aligned}
\]

Alternatively, we could have worked with the complementary angle, \(\tan 30^{\circ}=\frac{x}{6}\).

\section*{Exercise 12B}
1 Find the exact value of \(x\).

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-07(3)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-07(4)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-07}
\end{center}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-07(1)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-07(2)}
\end{center}

2 Find the exact values, rationalising the denominator where appropriate.
a \(\left(\sin 60^{\circ}\right)^{2}+\left(\cos 60^{\circ}\right)^{2}\)
b \(\left(\tan 30^{\circ}\right)^{2}-\frac{1}{\left(\cos 30^{\circ}\right)^{2}}\)
c \(\frac{\tan 60^{\circ}-\tan 45^{\circ}}{1+\tan 60^{\circ} \times \tan 45^{\circ}}\)
d \(\sin 45^{\circ} \times \cos 60^{\circ}+\cos 45^{\circ} \times \sin 60^{\circ}\)
e \(2 \sin 30^{\circ} \times \cos 30^{\circ}\)
f \(2\left(\cos 45^{\circ}\right)^{2}-1\)
\(3 A B C D\) is a rhombus with \(\angle A B D=30^{\circ}\). Find the exact length of each diagonal if the side lengths are \(10 \mathrm{~cm}\).

4 Find exact values of:
a \(A C\)
b \(A D\)
c \(B C\)
d \(D C\)
e \(B D\)

5 Find the exact values of \(a\) and \(x\).
\includegraphics[max width=\textwidth, center]{2024_02_22_e203939eb913b173edc2g-08(3)}

6 Find the exact value of \(x\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-08}
\end{center}

\(7 \quad A B C D\) is a rhombus with sides \(10 \mathrm{~cm}\). Find \(A X\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-08(1)}
\end{center}

\section*{2 
 Three-dimensional trigonometry}
We can apply our knowledge of trigonometry to solve problems in three dimensions. To do this you will need to draw careful diagrams and look for right-angled triangles. Sometimes it is helpful to draw a separate diagram showing the right-angled triangle.

\section*{Example 5}
In the triangular prism shown, find:

a the length \(C F\)

b the length \(B F\)

c the angle \(B F C\), correct to one decimal place.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-08(2)}
\end{center}

\section*{Solution}
a Applying Pythagoras' theorem to \(\triangle C E F\) :

\[
\begin{aligned}
C F^{2} & =4^{2}+5^{2} \\
& =41
\end{aligned}
\]

Hence, \(C F=\sqrt{41} \mathrm{~cm}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-09}
\end{center}

b Applying Pythagoras' theorem to \(\triangle B C F\) :

\[
\begin{aligned}
B F^{2} & =3^{2}+(\sqrt{41})^{2} \\
& =50
\end{aligned}
\]

Hence, \(B F=5 \sqrt{2} \mathrm{~cm}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-09(4)}
\end{center}

c To find the angle \(B F C\), draw \(\triangle B C F\) and let \(\angle B F C=\theta\).

Now \(\tan \theta=\frac{3}{\sqrt{41}}\)

so \(\quad \theta \approx 25.1^{\circ}\)

(Correct to one decimal place.)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-09(1)}
\end{center}

\section*{Angles of elevation and depression}
When a person looks at an object that is higher than the person's eye, the angle between the line of sight and the horizontal is called the angle of elevation.

On the other hand, when the object is lower than the person's eye, the angle between the horizontal and the line of sight is

\includegraphics[max width=\textwidth, center]{2024_02_22_e203939eb913b173edc2g-09(3)}
called the angle of depression.

In practice, 'eye of observer' is replaced by a point on the ground.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-09(2)}
\end{center}

\section*{Bearings}
Bearings are used to indicate the direction of an object from a fixed reference point, \(O\). True bearings give the angle \(\theta^{\circ}\) from north, measured clockwise. We write a true bearing of \(\theta^{\circ}\) as \(\theta^{\circ} \mathrm{T}\), where \(\theta^{\circ}\) is an angle between \(0^{\circ}\) and \(360^{\circ}\). It is customary to write the angle using three digits, so \(0^{\circ} \mathrm{T}\) is written \(000^{\circ} \mathrm{T}, 15^{\circ} \mathrm{T}\) is written \(015^{\circ} \mathrm{T}\), and so on.

For example, in the diagram opposite, the true bearing of \(A\) from \(O\) is \(060^{\circ} \mathrm{T}\), and the true bearing of \(B\) from \(O\) is \(140^{\circ} \mathrm{T}\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-09(5)}
\end{center}

\section*{Example 6}
A tower is situated due north of a point \(A\) and due west of a point \(B\). From \(A\), the angle of elevation of the top of the tower is \(18^{\circ}\). In addition, \(B\) (which is on the same level as \(A\) ) is 52 metres from \(A\) and has a bearing of \(064^{\circ} \mathrm{T}\) from \(A\). Find, correct to one decimal place:

a the distance from \(A\) to the base of the tower

b the height of the tower

c the angle of elevation of the top of the tower from \(B\).

\section*{Solution}
Draw the tower \(O T\) and mark the point \(A\) level with the base of the tower. The line \(A O\) then points north. We can then mark all the given information on the diagram. The triangle \(A O T\) is vertical and triangle \(A O B\) is horizontal.

a \(\operatorname{In} \triangle A O B, \cos 64^{\circ}=\frac{O A}{52}\)

So

\[
\begin{aligned}
O A & =52 \cos 64^{\circ} \\
& =22.795 \ldots \quad(\text { Keep this in your } \\
& \quad \text { calculator for part b.) }
\end{aligned}
\]

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-10(1)}
\end{center}

\(\approx 22.8 \quad\) (Correct to one decimal place.)

\(A\) is approximately 22.8 metres from the base of the tower.

b In \(\triangle A O T, \tan 18^{\circ}=\frac{O T}{O A}\)

that is,

\[
\begin{aligned}
O T & =O A \times \tan 18^{\circ} \\
& =7.406 \ldots \quad(\text { Keep this in your calculator for part } \mathbf{c} .) \\
& \approx 7.4 \quad \quad \text { (Correct to one decimal place. })
\end{aligned}
\]

The tower is approximately 7.4 metres high.

c In \(\triangle T O B, \tan \theta=\frac{O T}{O B}\)

Now from \(\triangle A O B, O B=52 \sin 64^{\circ}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-10}
\end{center}

Hence, \(\tan \theta=\frac{O T}{52 \sin 64^{\circ}}\)

\[
\approx 0.1585
\]

so

\[
\theta \approx 9.0^{\circ} \quad \text { (Correct to one decimal place.) }
\]

The angle of elevation of the top of the tower from \(B\) is approximately \(9.0^{\circ}\).

Do not re-enter a rounded result into your calculator; it is much more accurate to store the un-rounded number and use it in subsequent steps.

\section*{Exercise \(12 \mathrm{C}\)}
1 In the rectangular prism shown opposite, find:

a \(B N\)

b \(\angle B N M\) (correct to one decimal place)

c \(B P\)

d the angle BPM (correct to one decimal place)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-11(3)}
\end{center}

e \(M Q\), where \(Q\) is the midpoint of \(P N\)

f the angle \(B Q M\) (correct to one decimal place)

2 In the cube shown opposite, find:

a \(C E\)

b \(\angle C E G\) (correct to one decimal place)

c \(\angle C B E\)

d \(\angle C E B\) (correct to one decimal place)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-11(2)}
\end{center}

3 In the square pyramid shown opposite, find:
a \(A C\)
b \(O C\)
c \(V C\)

d \(\angle V C O\) (correct to one decimal place)

e \(O M\), where \(M\) is the midpoint of \(B C\)

f \(\angle V M O\) (correct to one decimal place)

g \(\angle V B M\) (correct to one decimal place)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-11}
\end{center}

\(4 A E F D\) is a horizontal rectangle. \(A B C D\) is a rectangle inclined at an angle \(\theta\) to the horizontal. \(A D=32 \mathrm{~cm}, A E=24 \mathrm{~cm}\) and \(B E=41 \mathrm{~cm}\). Find, correct to one decimal place where necessary:

a \(D C\)

b \(A F\)

c \(\angle C A F\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-11(1)}
\end{center}

5 The base of a tree is situated 50 metres due north of a point \(P\). The angle of elevation of the top of the tree from \(P\) is \(32^{\circ}\).

a Find the height of the tree, correct to one decimal place.

b \(Q\) is a point 100 metres due east of \(P\). Find:

i the distance of \(Q\) from the base of the tree

ii the angle of elevation of the top of the tree from \(Q\), correct to one decimal place

iii the bearing of the tree from \(Q\), correct to one decimal place

6 Dillon and Eugene are both looking at a tower of height 35 metres. Dillon is standing due south of the tower and he measures the angle of elevation from the ground to the top of the tower to be \(15^{\circ}\). Eugene is standing due east of the tower and he measures the angle of elevation from the ground to the top of the tower to be \(20^{\circ}\). Find, correct to one decimal place:

a the distance Dillon is from the foot of the tower

b the distance Eugene is from the foot of the tower

c the distance between Dillon and Eugene

d the bearing of Dillon from Eugene

7 From a point \(A\), a lighthouse is on a bearing of \(026^{\circ} \mathrm{T}\) and the top of the lighthouse is at angle of elevation of \(20.25^{\circ}\). From a point \(B\), the lighthouse is on a bearing of \(296^{\circ} \mathrm{T}\) and the top of the lighthouse is at angle of elevation of \(10.20^{\circ}\). If \(A\) and \(B\) are 500 metres apart, find the height of the lighthouse, correct to the nearest metre.

8 From the top of a cliff that runs north-south, the angle of depression of a yacht, 200 metres out to sea and due east of the observer, is \(20^{\circ}\). When the observer next looks at the yacht, he notices that it has sailed 150 metres parallel to the cliff.

a Find the height of the cliff, correct to the nearest metre.

b Find the distance the yacht is from the observer after it has sailed 150 metres parallel to the cliff, correct to the nearest metre.

c Find the angle of depression of the yacht from the top of the cliff when it is in its new position, correct to the nearest degree.

9 A mast is held in position by means of two taut ropes running from the ground to the top of the mast. One rope is of length 40 metres and makes an angle of \(58^{\circ}\) with the ground. Its anchor point with the ground is due south of the mast. The other rope is 50 metres long and its anchor point is due east of the mast. Find the distance, correct to the nearest metre, between the two anchor points.

\section*{The sine rule}
In many situations we encounter triangles that are not right-angled. We can use trigonometry to deal with these triangles as well. One of the two key formulas for doing this is known as the sine rule.

We begin with an acute-angled triangle, \(A B C\), with side lengths \(a, b\) and \(c\), as shown below. (It is standard to write a lower case letter on a side and the corresponding upper case letter on the angle opposite that side.) Drop a perpendicular, \(C P\), of length \(h\), from \(C\) to \(A B\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-13}
\end{center}

In \(\triangle A P C\) we have \(\sin A=\frac{h}{b}\), so \(h=b \sin A\).

Similarly, in \(\triangle C P B\) we have \(\sin B=\frac{h}{a}\), so \(h=a \sin B\).

Equating these expressions for \(h\), we have:

\[
b \sin A=a \sin B
\]

which we can write as:

\[
\frac{a}{\sin A}=\frac{b}{\sin B}
\]

The same result holds for the side \(c\) and angle \(C\), so we can write:

\[
\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}
\]

This is known as the sine rule.

In words, this says 'any side of a triangle over the sine of the opposite angle equals any other side of the triangle over the sine of its opposite angle'.

This result also holds in an obtuse-angled triangle. We will look at that case later.

\section*{The sine rule}
In any triangle \(A B C\) :

\(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-13(1)}
\end{center}

For example, the sine rule can be used to find an unknown length of a side of a triangle when a side length and the angles are known. This is closely related to the AAS congruence test.

\section*{Example 7}
In \(\triangle A B C=A B=9 \mathrm{~cm}, \angle A B C=76^{\circ}\) and \(\angle A C B=58^{\circ}\).

Find, correct to two decimal places:
a \(A C\)
b \(B C\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-14}
\end{center}

\section*{Solution}
a Apply the sine rule:

\[
\begin{aligned}
\frac{A C}{\sin 76^{\circ}} & =\frac{9}{\sin 58^{\circ}} \\
A C & =\frac{9 \sin 76^{\circ}}{\sin 58^{\circ}} \\
& \approx 10.30 \mathrm{~cm}
\end{aligned}
\]

b To find \(B C\), we need the angle \(\angle C A B\) opposite it.

\[
\begin{aligned}
\angle C A B & =180^{\circ}-58^{\circ}-76^{\circ} \\
& =46^{\circ}
\end{aligned}
\]

Then by the sine rule:

\[
\begin{aligned}
\frac{B C}{\sin 46^{\circ}} & =\frac{9}{\sin 58^{\circ}} \\
B C & =\frac{9 \sin 46^{\circ}}{\sin 58^{\circ}} \\
& \approx 7.63 \mathrm{~cm}
\end{aligned}
\]

\section*{Example 8}
From two points \(A\) and \(B\), which are 800 metres apart on a straight north-south road, the bearings of a house are \(125^{\circ} \mathrm{T}\) and \(050^{\circ} \mathrm{T}\), respectively. Find how far each point is from the house, correct to the nearest metre.

\section*{Solution}
We draw a diagram to represent the information.

We can find the angles in \(\triangle A H B\).

\[
\begin{aligned}
\angle H A B & =180^{\circ}-125^{\circ} \\
& =55^{\circ}
\end{aligned}
\]

and \(\angle A H B=180^{\circ}-50^{\circ}-55^{\circ}\)

\[
=75^{\circ}
\]

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-14(1)}
\end{center}

Apply the sine rule to \(\triangle A B H\) :

\[
\begin{aligned}
\frac{B H}{\sin 55^{\circ}} & =\frac{800}{\sin 75^{\circ}} \\
B H & =\frac{800 \sin 55^{\circ}}{\sin 75^{\circ}} \\
& \approx 678.44 \mathrm{~m}
\end{aligned}
\]

\[
\text { Similarly, } \frac{A H}{\sin 50^{\circ}}=\frac{800}{\sin 75^{\circ}}
\]

\[
\begin{aligned}
A H & =\frac{800 \sin 50^{\circ}}{\sin 75^{\circ}} \\
& \approx 634.45 \mathrm{~m}
\end{aligned}
\]

Thus, \(A\) and \(B\) are approximately 634 metres and 678 metres from the house, respectively.

\section*{Finding angles}
The sine rule can also be used to find angles in a triangle, provided that one of the known sides is opposite a known angle.

At this stage we can only deal with acute angled triangles.

\section*{Example 9}
Find the angle \(\theta\) in the triangle \(F G H\), correct to the nearest degree.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-15}
\end{center}

\section*{Solution}
Apply the sine rule to \(\triangle F G H\) :

\(\frac{8}{\sin \theta}=\frac{12}{\sin 75^{\circ}}\)

To make the algebra easier, take the reciprocal of both sides:

\[
\frac{\sin \theta}{8}=\frac{\sin 75^{\circ}}{12}
\]

Hence, \(\sin \theta=\frac{8 \sin 75^{\circ}}{12}\)

\[
=0.6440 \ldots
\]

Hence, \(\theta \approx 40^{\circ}\) (Correct to the nearest degree.)

\section*{Example 10}
Find the length of \(O C\) in the diagram, correct to one decimal place.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-15(1)}
\end{center}

\section*{Solution}
\(O C=h \mathrm{~m}\) and \(B C=x \mathrm{~m}\).

\(\angle A C B+32^{\circ}=70^{\circ}\) (exterior angle of \(\triangle A B C\) )

The angle \(\angle A C B=38^{\circ}\)

Applying the sine rule:

\[
\begin{aligned}
\frac{x}{\sin 32^{\circ}} & =\frac{12}{\sin 38^{\circ}} \\
x & =\frac{12 \sin 32^{\circ}}{\sin 38^{\circ}} \\
& =10.3287 \ldots \quad \text { (Keep this in your calculator.) }
\end{aligned}
\]

In triangle \(B C O\) :

\[
\sin 70^{\circ}=\frac{h}{x}
\]

so

\[
\begin{aligned}
h & =x \times \sin 70^{\circ} \\
& \approx 9.7 \quad \text { (Correct to one decimal place.) }
\end{aligned}
\]

The length \(O C\) is approximately \(9.7 \mathrm{~m}\).

Note: Alternatively, \(h\) can be calculated directly as \(\frac{12 \sin 32^{\circ}}{\sin 38^{\circ}} \times \sin 70^{\circ}\).

\section*{Exercise 12D}
1 Find the value of \(b\), correct to two decimal places.
\includegraphics[max width=\textwidth, center]{2024_02_22_e203939eb913b173edc2g-16(3)}

2 Find the value of \(x\), correct to two decimal places.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-16(1)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-16(2)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-16}
\end{center}

3 a In \(\triangle A B C, A=62^{\circ}, B=54^{\circ}\) and \(a=8\). Find \(b\), correct to two decimal places.

b In \(\triangle A B C, B=47^{\circ}, C=82^{\circ}\) and \(b=10\). Find \(a\), correct to two decimal places.

4 Find the value of \(\theta\), correct to the nearest degree.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-17}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-17(5)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-17(3)}
\end{center}

5 In \(\triangle A B C, A=71^{\circ}, a=18 \mathrm{~cm}\) and \(b=14 \mathrm{~cm}\). Find, correct to two decimal places:
a \(B\)
b \(C\)
c \(c\)

\(6 A B C D\) is a parallelogram with \(\angle A D C=50^{\circ}\). The shorter diagonal, \(A C\), is \(20 \mathrm{~m}\), and \(A D=15 \mathrm{~m}\). Find \(\angle A C D\) and hence the length of the side \(D C\), correct to two decimal places.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-17(1)}
\end{center}

7 Two hikers, Paul and Sayo, are both looking at a distant landmark. From Paul, the bearing of the landmark is \(222^{\circ} \mathrm{T}\) and, from Sayo, the bearing of the landmark is \(300^{\circ} \mathrm{T}\). If Sayo is standing \(800 \mathrm{~m}\) due south of Paul, find, correct to the nearest metre:

a the distance from Paul to the landmark

b the distance from Sayo to the landmark

8 An archaeologist wishes to determine the height of an ancient temple. From a point \(A\) at ground level, she measures the angle of elevation of \(V\), the top of the temple, to be \(37^{\circ}\).

She then walks \(100 \mathrm{~m}\) towards the temple to a point \(B\). From here, the angle of elevation of \(V\) from ground level is \(64^{\circ}\). Find:

a \(\angle A V B\)

b \(V B\), correct to two decimal places

c \(O V\), the height of the temple, to the nearest metre

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-17(4)}
\end{center}

9 A hillside is inclined at \(26^{\circ}\) to the horizontal. From the bottom of the hill, Alex observes a vertical tree whose base is \(40 \mathrm{~m}\) up the hill from the point where Alex is standing. If the angle of elevation of the top of the tree is \(43^{\circ}\) from the point where Alex is standing, find the height of the tree, correct to the nearest metre.

10 Find \(h\), correct to the nearest centimetre.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-17(2)}
\end{center}

\section*{12 Trigonometric ratios of obtuse angles}
We have seen that we can use the sine rule to find sides and angles in acute-angled triangles. What happens when one of the angles is obtuse? We can extend our definition of the basic trigonometric functions to obtuse angles by using coordinate geometry.

We begin by drawing a circle of radius 1 in the Cartesian plane, with its centre at the origin. The equation of the circle is \(x^{2}+y^{2}=1\).

Take a point \(P(a, b)\) on the circle in the first quadrant and form the right-angled triangle \(P O Q\) with \(O\) at the origin. Let \(\angle P O Q\) be \(\theta\).

\[
\begin{aligned}
& \cos \theta=\frac{O Q}{O P}=\frac{a}{1}=a, \text { and } \\
& \sin \theta=\frac{Q P}{O P}=\frac{b}{1}=b
\end{aligned}
\]

But \(a\) is the \(x\)-coordinate of \(P\) and \(b\) is the \(y\)-coordinate of \(P\).

Hence, the coordinates of the point \(P\) are \((\cos \theta, \sin \theta)\).

We can now turn this idea around and say that if \(\theta\) is the angle between \(O P\) and the positive \(x\)-axis, then:

\begin{itemize}
  \item the cosine of \(\theta\) is defined to be the \(x\)-coordinate of the point \(P\) on the unit circle
  \item the sine of \(\theta\) is defined to be the \(y\)-coordinate of the point \(P\) on the unit circle.
\end{itemize}

This definition can be applied to all angles \(\theta\), but in this chapter we will restrict the angle \(\theta\) to \(0^{\circ} \leq \theta \leq 180^{\circ}\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-18}
\end{center}

Now take \(\theta\) to be \(30^{\circ}\), so \(P\) has coordinates \(\left(\cos 30^{\circ}, \sin 30^{\circ}\right)\).

Suppose that we move the point \(P\) around the circle to \(P^{\prime}\) so that \(P^{\prime}\) makes an angle of \(150^{\circ}\) with the positive \(x\)-axis. (Recall that \(30^{\circ}\) and \(150^{\circ}\) are supplementary angles.)

The coordinates of \(P^{\prime}\) are \(\left(\cos 150^{\circ}, \sin 150^{\circ}\right)\). But we can see that triangles \(O P Q\) and \(O P^{\prime} Q^{\prime}\) are congruent, so the \(y\)-coordinates of \(P\) and \(P^{\prime}\) are the same. That is:

\[
\sin 150^{\circ}=\sin 30^{\circ}
\]

The \(x\)-coordinates have the same magnitude but opposite sign, so:

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-18(1)}
\end{center}

\[
\cos 150^{\circ}=-\cos 30^{\circ}
\]

From this example, we can see the following rules.

\section*{Supplementary angles}
\begin{itemize}
  \item The sines of two supplementary angles are the same.
  \item The cosines of two supplementary angles are opposite in sign.
  \item In symbols:
\end{itemize}

\(\sin \theta=\sin \left(180^{\circ}-\theta\right)\) and \(\cos \theta=-\cos \left(180^{\circ}-\theta\right)\)

We can extend the definition of sine and cosine to angles beyond \(180^{\circ}\). This will be done later in this book.

\section*{The angles \(0^{\circ}, 90^{\circ}\) and \(180^{\circ}\)}
We have defined \(\cos \theta\) and \(\sin \theta\) as the \(x\) - and \(y\)-coordinates of the point \(P\) on the unit circle. Taking the axis intercepts \(P_{1}, P_{2}\) and \(P_{3}\) from the unit circle diagram to the right, we obtain the following table of values. These values should be memorised.

\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
\(\theta\) & \(0^{\circ}\) & \(90^{\circ}\) & \(180^{\circ}\) \\
\hline
\(\sin \theta\) & 0 & 1 & 0 \\
\hline
\(\cos \theta\) & 1 & 0 & -1 \\
\hline
\end{tabular}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-19}
\end{center}

\section*{Example 11}
Find the exact value of:
a \(\sin 150^{\circ}\)
b \(\cos 150^{\circ}\)
c \(\sin 120^{\circ}\)
d \(\cos 120^{\circ}\)

\section*{Solution}
a \(\sin 150^{\circ}=\sin (180-150)^{\circ}\)

\(=\sin 30^{\circ}\)

\(=\frac{1}{2}\)

c \(\sin 120^{\circ}=\sin (180-120)^{\circ}\)

\(=\sin 60^{\circ}\)

\(=\frac{\sqrt{3}}{2}\) b \(\cos 150^{\circ}=-\cos (180-150)^{\circ}\)

\(=-\cos 30^{\circ}\)

\(=-\frac{\sqrt{3}}{2}\)

d \(\cos 120^{\circ}=-\cos (180-120)^{\circ}\)

\(=-\cos 60^{\circ}\)

\(=-\frac{1}{2}\)

Note: You can verify these results using your calculator.

\section*{Example 12}
Find, correct to the nearest degree, the acute and obtuse angle whose sine is:
a approximately 0.7431
b \(\frac{1}{\sqrt{2}}\)
c \(\frac{\sqrt{3}}{2}\)

\section*{Solution}
a If \(\sin \theta=0.7431\) and \(\theta\) is acute, then the calculator gives \(\theta=\sin ^{-1} 0.7431 \approx 48^{\circ}\). Hence, the solutions are \(48^{\circ}\) and \(132^{\circ}\), correct to the nearest degree, because \(132^{\circ}\) is the supplement of \(48^{\circ}\).

b If \(\sin \theta=\frac{1}{\sqrt{2}}\)

\[
\theta=45^{\circ} \text { or } \theta=180^{\circ}-45^{\circ}
\]

That is, \(\theta=45^{\circ}\) or \(\theta=135^{\circ}\)

c If \(\sin \theta=\frac{\sqrt{3}}{2}\)

\[
\theta=60^{\circ} \text { or } \theta=180^{\circ}-60^{\circ}
\]

That is, \(\theta=60^{\circ}\) or \(\theta=120^{\circ}\)

\section*{More on the sine rule}
The sine rule also holds in obtuse-angled triangles. A proof is given in question 7 of Exercise 12E. We now see how to apply the sine rule in obtuse-angled triangles.

\section*{Example 13}
Find the value of \(x\), correct to one decimal place.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-20}
\end{center}

\section*{Solution}
Apply the sine rule to \(\triangle A B C\) :

\[
\begin{aligned}
\frac{x}{\sin 130^{\circ}} & =\frac{7}{\sin 20^{\circ}} \\
x & =\frac{7 \sin 130^{\circ}}{\sin 20^{\circ}}\left(\sin 130^{\circ}=\sin 50^{\circ}\right) \\
& \approx 15.7 \quad(\text { Correct to one decimal place. })
\end{aligned}
\]

\section*{The ambiguous case}
You are given the following information about a triangle. A triangle has side lengths \(9 \mathrm{~m}\) and \(7 \mathrm{~m}\) and an angle of \(45^{\circ}\) between the \(9 \mathrm{~m}\) side and the unknown side. How many triangles satisfy these properties?

In the diagram, the triangles \(A B C\) and \(A B C^{\prime}\) both have sides of length \(9 \mathrm{~m}\) and \(7 \mathrm{~m}\), and both contain an angle of \(45^{\circ}\) opposite the side of length \(7 \mathrm{~m}\). Despite this, the triangles are

\includegraphics[max width=\textwidth, center]{2024_02_22_e203939eb913b173edc2g-21}
different. (Recall that the included angle was required in the SAS congruence test.)

Hence, given the data that a triangle \(\angle P Q R\) has \(P Q=9 \mathrm{~m}\), \(\angle P Q R=45^{\circ}\) and \(P R=7 \mathrm{~m}\), the angle opposite \(P Q\) is not determined. There are two non-congruent triangles that satisfy the given data.

Let \(P R^{\prime}=7 \mathrm{~m}\) so that \(\theta=\angle P R Q\) is acute and \(\theta^{\prime}=\angle P R^{\prime} Q\) is obtuse.

Applying the sine rule to the \(\triangle P R Q\), we have:

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-21(1)}
\end{center}

\[
\begin{aligned}
\frac{9}{\sin \theta} & =\frac{7}{\sin 45^{\circ}} \\
\sin \theta & =\frac{9 \sin 45^{\circ}}{7} \\
& =0.9091 \ldots
\end{aligned}
\]

The calculator tells us that \(\sin ^{-1}(0.9091 \ldots)\) is approximately \(65^{\circ}\). Hence \(\theta \approx 65^{\circ}\).

The triangle \(P R^{\prime} R\) is isosceles, so \(\angle P R^{\prime} R\) is \(65^{\circ}\) and \(\theta^{\prime}=180^{\circ}-65^{\circ}=115^{\circ}\). Since \(\sin 65^{\circ}=\sin 115^{\circ}\), the triangle \(P R^{\prime} Q\) also satisfies the given data.

\section*{Exercise 12E}
1 Copy and complete:
a \(\sin 115^{\circ}=\sin\)
b \(\cos 123^{\circ}=-\cos\)
c \(\sin 138^{\circ}=\sin\)
d \(\cos 95^{\circ}=-\cos\)

2 Find the exact value of:
a \(\sin 135^{\circ}\)
b \(\cos 135^{\circ}\)

3 a Find the acute and the obtuse angle whose sine is \(\frac{1}{2}\).

b Find, correct to the nearest degree, two angles whose sine is approximately 0.5738 .

c Find, correct to the nearest degree, an angle whose cosine is approximately -0.8746 .

4 Copy and complete:

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
\(\theta\) & \(30^{\circ}\) & \(120^{\circ}\) & \(150^{\circ}\) & \(90^{\circ}\) & \(135^{\circ}\) \\
\hline
\(\sin \theta\) &  & \(\frac{\sqrt{3}}{2}\) &  &  & \(\frac{1}{\sqrt{2}}\) \\
\hline
\(\cos \theta\) &  &  & \(-\frac{\sqrt{3}}{2}\) & 0 &  \\
\hline
\end{tabular}
\end{center}

5 Use the sine rule to find the value of \(x\), correct to two decimal places.
\includegraphics[max width=\textwidth, center]{2024_02_22_e203939eb913b173edc2g-22(4)}

6 Given that \(\theta\) is an obtuse angle, find its value, correct to the nearest degree.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-22(1)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-22}
\end{center}

7 Suppose that \(\angle A\) in triangle \(A B C\) is obtuse.

a Explain why \(\sin \angle A=\sin \angle C A M\).

b Use triangle \(A C M\) to find a formula for \(h\) in terms of \(A\) and \(b\).

c Use triangle \(B C M\) to find a formula for \(h\) in terms of \(B\) and \(a\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-22(2)}
\end{center}

d Deduce that \(\frac{a}{\sin A}=\frac{b}{\sin B}\).

That is, we have proved the sine rule holds in obtuse-angled triangles.

8 The angle between the two sides of a parallelogram is \(93^{\circ}\). If the longer side has length \(12 \mathrm{~cm}\) and the longer diagonal has length \(14 \mathrm{~cm}\), find the angle between the long diagonal and the short side of the parallelogram, correct to the nearest degree.

\(9 A B C D\) is a parallelogram. \(\angle C D A=130^{\circ}\), the long diagonal \(A C\) is \(50 \mathrm{~m}\) and \(A D=30 \mathrm{~m}\). Find the length of the side \(D C\), correct to one decimal place.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-22(3)}
\end{center}

10 Sonia starts at \(O\) and walks 600 metres due east to point \(A\). She then walks on a bearing of \(250^{\circ} \mathrm{T}\) to point \(B, 750\) metres from \(O\). Find:

a the bearing of \(B\) from \(O\), correct to the nearest degree

b the distance from \(A\) to \(B\), correct to the nearest metre

11 A point \(M\) is one kilometre due east of a point \(C\). A hill is on a bearing of \(028^{\circ} \mathrm{T}\) from \(C\) and is \(1.2 \mathrm{~km}\) from \(M\). Find:

a the bearing of the hill from \(M\), correct to the nearest degree

b the distance, correct to the nearest metre, between \(C\) and the hill

\section*{12 The cosine rule}
We know, from the SAS congruence test, that a triangle is completely determined if we are given two of its sides and the included angle. If we want to know the third side and the two other angles, the sine rule does not help us.

You can see from the diagram that there is not enough information to apply the sine rule. This is because the known angle is not opposite one of the known sides.

Fortunately there is another rule called the cosine rule which we can use in this situation.

Suppose that \(A B C\) is a triangle and that the angles \(A\) and \(C\) are acute. Drop a perpendicular from \(B\) to \(A C\) and mark the side lengths as shown in the diagram.

In \(\triangle B D A\), Pythagoras' theorem tells us that:

\[
c^{2}=h^{2}+(b-x)^{2}
\]

Also in \(\triangle C B D\), by Pythagoras' theorem we have:

\[
h^{2}=a^{2}-x^{2}
\]

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-23}
\end{center}

Substituting this expression for \(h^{2}\) into the first equation and expanding:

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-23(1)}
\end{center}

\[
\begin{aligned}
c^{2} & =a^{2}-x^{2}(b-x)^{2} \\
& =a^{2}-x^{2}+b^{2}-2 b x+x^{2} \\
& =a^{2}+b^{2}-2 b x
\end{aligned}
\]

Finally, from \(\triangle C B D\), we have \(\frac{x}{a}=\cos C\). That is, \(x=a \cos C\) and so:

\[
c^{2}=a^{2}+b^{2}-2 a b \cos C
\]

\section*{Notes:}
\begin{itemize}
  \item By relabelling the sides and angle, we could also write \(a^{2}=b^{2}+c^{2}-2 b c \cos A\) and \(b^{2}=a^{2}+c^{2}-2 a c \cos B\).
  \item If \(C=90^{\circ}\), then, since \(\cos 90^{\circ}=0\), we obtain Pythagoras' theorem. Thus the cosine rule can be thought of as 'Pythagoras' theorem with a correction term'.
  \item The cosine rule is also true if \(C\) is obtuse. This is proven in the exercises.
\end{itemize}

\section*{Example 14}
Find the value of \(x\), correct to one decimal place.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-24}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-24(2)}
\end{center}

\section*{Solution}
a Applying the cosine rule to \(\triangle A B C\) :

\[
\begin{aligned}
x^{2} & =10^{2}+15^{2}-2 \times 10 \times 15 \times \cos 50^{\circ} \\
& =132.16 \ldots
\end{aligned}
\]

so \(x \approx 11.5\) b Applying the cosine rule:

\[
\begin{aligned}
x^{2} & =7^{2}+8^{2}-2 \times 7 \times 8 \times \cos 110^{\circ} \\
& =151.30 \ldots
\end{aligned}
\]

so \(x \approx 12.3\) (Correct to one decimal place.)

Note that in Example 14a, \(x^{2}<10^{2}+15^{2}\) since \(\cos 50^{\circ}\) is positive. In Example 14b, \(x^{2}>7^{2}+8^{2}\) since \(\cos 110^{\circ}\) is negative.

\section*{Example 15}
A tower at \(A\) is 450 metres from \(O\) on a bearing of \(340^{\circ} \mathrm{T}\) and a tower at \(B\) is 600 metres from \(O\) on a bearing of \(060^{\circ} \mathrm{T}\). Find, correct to the nearest metre, the distance between the two towers.

\section*{Solution}
We draw a diagram to represent the information.

Now \(\angle A O B=80^{\circ}\). Let \(A B=x \mathrm{~m}\).

Applying the cosine rule:

\[
\begin{aligned}
x^{2} & =450^{2}+600^{2}-2 \times 450 \times 600 \times \cos 80^{\circ} \\
& =468729.98 \ldots
\end{aligned}
\]

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-24(1)}
\end{center}

that is, \(x \approx 684.63 \ldots\)

Hence, the towers are 685 metres apart, correct to the nearest metre.

\section*{The cosine rule}
In any triangle \(A B C\) :

\(a^{2}=b^{2}+c^{2}-2 b c \cos A\),

where \(A\) is the angle opposite \(a\).

The cosine rule can be used to find the length of the third side of a triangle when the lengths of two sides and the

\includegraphics[max width=\textwidth, center]{2024_02_22_e203939eb913b173edc2g-25}
size of the included angle are known.

\section*{Exercise 12F}
1 In each triangle, calculate the unknown side length, giving your answer correct to two decimal places.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-25(4)}
\end{center}

\(\mathbf{c}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-25(2)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-25(3)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-25(1)}
\end{center}

\(2 A B C D\) is a parallelogram with sides \(15 \mathrm{~cm}\) and \(18 \mathrm{~cm}\). The angle at \(A\) is \(65^{\circ}\). Find the length of the shorter diagonal, correct to two decimal places.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-25(6)}
\end{center}

3 A vertical pole \(O V\) is being held in position by two ropes, \(V A\) and \(V B\). If \(V A=6 \mathrm{~m}, V B=6.5 \mathrm{~m}, \angle O V B=32^{\circ}\) and \(\angle O V A=27^{\circ}\) find, correct to one decimal place, the distance \(A B\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-25(5)}
\end{center}

4 A ship is \(300 \mathrm{~km}\) from port on a bearing of \(070^{\circ} \mathrm{T}\). A second ship is \(400 \mathrm{~km}\) from the same port and on a bearing of \(140^{\circ} \mathrm{T}\). How far apart, correct to the nearest kilometre, are the two ships?

5 A pilot flies a plane on course for an airport \(600 \mathrm{~km}\) away. Unfortunately, due to an error, his bearing is out by \(2^{\circ}\). After travelling \(700 \mathrm{~km}\) he realises he is off course. How far from the airport is he, correct to the nearest kilometre?

6 A rhombus \(P Q R S\) has side lengths \(8 \mathrm{~m}\), and contains an angle of \(128^{\circ}\).

a Find the length of the longer diagonal, correct to two decimal places.

b Find the length of the shorter diagonal, correct to two decimal places.

c Find the area of the rhombus, correct to two decimal places.

7 Prove the cosine rule when the included angle, \(A\) is obtuse.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-26(2)}
\end{center}

\section*{2 Finding angles using the cosine rule}
The SSS congruence test tells us that once three sides of a triangle are known, the angles are uniquely determined. The question is, how do we find them?

Given three sides of a triangle, we can substitute the information into the cosine rule and rearrange to find the cosine of one of the angles and hence the angle.

If you prefer, you can learn or derive another form of the cosine rule, with \(\cos C\) as the subject.

Rearranging \(c^{2}=a^{2}+b^{2}-2 a b \cos C\) we have:

\(2 a b \cos C=a^{2}+b^{2}-c^{2}\)

\(\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-26}
\end{center}

\section*{Example 16}
A triangle has side lengths \(6 \mathrm{~cm}, 8 \mathrm{~cm}\) and \(11 \mathrm{~cm}\). Find the smallest angle in the triangle.

\section*{Solution}
The smallest angle in the triangle is opposite the smallest side.

Applying the cosine rule:

\[
\begin{aligned}
6^{2} & =8^{2}+11^{2}-2 \times 8 \times 11 \times \cos \theta \\
\cos \theta & =\frac{8^{2}+11^{2}-6^{2}}{2 \times 8 \times 11} \\
& =\frac{149}{172}
\end{aligned}
\]

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-26(1)}
\end{center}

and so \(\theta \approx 32.2^{\circ} \quad\) (Correct to one decimal place.)

There is no ambiguous case when we use the cosine rule to find an angle. In the following example, the unknown angle is obtuse.

\section*{Example 17}
In \(A B C, a=6, b=20\) and \(c=17\). Find the size of \(\angle A B C\), correct to one decimal place.

\section*{Solution}
Applying the cosine rule:

\[
\begin{aligned}
20^{2} & =6^{2}+17^{2}-2 \times 6 \times 17 \cos B \\
\cos B & =\frac{6^{2}+17^{2}-20^{2}}{2 \times 6 \times 17} \\
& =\frac{-75}{204}
\end{aligned}
\]

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-27(1)}
\end{center}

and so \(B=111.6^{\circ}\)

(Correct to one decimal place.)

\section*{Using the cosine rule to find an angle}
\begin{itemize}
  \item The cosine rule can be used to determine the size of any angle in a triangle where the three side lengths are known.
  \item In any triangle \(\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}\), where \(C\) is the opposite angle \(C\).
\end{itemize}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-27}
\end{center}

\section*{Exercise 12G}
1 Copy and complete the statement of the cosine rule.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-27(6)}
\end{center}

\(x^{2}=\) b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-27(2)}
\end{center}

\(b^{2}=\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-27(5)}
\end{center}

\(p^{2}=\)

2 Calculate \(\alpha\), giving the answer correct to one decimal place.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-27(3)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-27(4)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-27(7)}
\end{center}

3 Find all angles. Give answers correct to one decimal place.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-28(1)}
\end{center}

b \(A\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-28(3)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-28}
\end{center}

4 Calculate the size of the smallest angle of the triangle whose side lengths are \(30 \mathrm{~mm}\), \(70 \mathrm{~mm}\) and \(85 \mathrm{~mm}\). Give your answer correct to one decimal place.

5 A triangle has sides of length \(9 \mathrm{~cm}, 13 \mathrm{~cm}\) and \(18 \mathrm{~cm}\). Calculate the size of the largest angle, correct to one decimal place.

6 Find all the angles of a triangle whose sides are in the ratio \(4: 8: 11\), to the nearest degree.

7 A parallelogram has sides of length \(12 \mathrm{~cm}\) and \(18 \mathrm{~cm}\). The longer diagonal has length \(22 \mathrm{~cm}\). Find, correct to one decimal place, the size of the obtuse angle between the two sides.

8 In \(\triangle A B C, A B=6 \mathrm{~cm}, A C=10 \mathrm{~cm}, B C=14 \mathrm{~cm}\) and \(X\) is the midpoint of side \(B C\). a Find, correct to one decimal place:

\[
\text { i } \angle \text { ACB } \quad \text { ii the length } A X
\]

\(A X\) is called a median of the triangle. A median is the line segment from a vertex to the midpoint of the opposite side.

b Find the length of the other two medians, correct to one decimal place.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-28(4)}
\end{center}

\section*{Area of a triangle}
If we know two sides and an included angle of a triangle, then by the SAS congruence test the area is determined. We will now find a formula for the area.

In \(\triangle A B C\) on the right, drop a perpendicular from \(A\) to \(B C\).

Then in \(\triangle A P C\) :

\[
\frac{h}{a}=\sin C
\]

That is:

\[
h=a \sin C
\]

Hence, the area of \(\triangle A B C=\frac{1}{2} b h=\frac{1}{2} a b \sin C\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-28(2)}
\end{center}

Thus, the area of a triangle is half the product of any two sides times the sine of the included angle.

\section*{Area of a triangle}
Area \(=\frac{1}{2} a b \sin C\), where \(C\) is included angle.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-29}
\end{center}

Note that if \(C=90^{\circ}\), then since \(\sin 90^{\circ}=1\), the area formula becomes \(\frac{1}{2} a b\). The formula also applies when the angle is obtuse. This is proved in the exercises.

\section*{Example 18}
Calculate the area of the triangle \(A B C\), correct to one decimal place.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-29(1)}
\end{center}

\section*{Solution}
Area of \(\triangle A B C=\frac{1}{2} \times 9 \times 15 \times \sin 42^{\circ}\)

\(\approx 45.2\) (Correct to one decimal place.)

So the area of the triangle is \(45.2 \mathrm{~cm}^{2}\).

\section*{Example 19}
The triangle shown has area \(34 \mathrm{~cm}^{2}\). Find the length of \(B C\), correct to two decimal places.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-29(2)}
\end{center}

\section*{Solution}
Let \(B C=x \mathrm{~cm}\)

\[
\begin{aligned}
34 & =\frac{1}{2} \times 7 \times x \times \sin 61^{\circ} \\
x & =\frac{68}{7 \sin 61^{\circ}} \\
& \approx 11.11 \quad \quad(\text { Correct to two decimal places.) } \\
B C & \approx 11.11 \mathrm{~cm}
\end{aligned}
\]

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-29(3)}
\end{center}

\section*{Exercise \(12 \mathrm{H}\)}
Example 181 Calculate each area, correct to one decimal place.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-30(6)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-30(9)}
\end{center}

\(\mathbf{c}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-30(1)}
\end{center}

2 Calculate each area, correct to two decimal places.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-30(8)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-30(7)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-30(4)}
\end{center}

3 In \(\triangle A B C\) shown opposite, \(\angle C A B\) is an acute angle.

a Use the sine rule to find \(\angle C A B\), correct to one decimal place.

b Find \(\angle A B C\), correct to one decimal place.

c Find the area of the triangle, correct to the nearest square centimetre.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-30(2)}
\end{center}

4 In \(\triangle A B C\) shown opposite:

a use the cosine rule to find \(\angle B A C\)

b find the area of the triangle, correct to the nearest square centimetre

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-30}
\end{center}

5 Calculate the area of each triangle, correct to the nearest square centimetre.
\includegraphics[max width=\textwidth, center]{2024_02_22_e203939eb913b173edc2g-30(10)}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-30(3)}
\end{center}

6 In \(\triangle A B C\) shown opposite, the area of the triangle is \(40 \mathrm{~cm}^{2}\).

Find, correct to one decimal place:
a \(A B\)
b \(A C\)
c \(\angle A C B\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-30(5)}
\end{center}

7 An acute-angled triangle of area \(60 \mathrm{~cm}^{2}\) has side lengths of \(16 \mathrm{~cm}\) and \(20 \mathrm{~cm}\). What is the magnitude of the included angle, correct to the nearest degree?

8 An irregular block of land, \(A B C D\), has dimensions shown opposite. Calculate, correct to one decimal place:

a the length \(A C\)

b \(\angle A B C\)

c the area of the block

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-31}
\end{center}

\(9 A B C D E\) is a regular pentagon with side lengths \(10 \mathrm{~cm}\). Diagonals \(A D\) and \(A C\) are drawn. Find:

a \(\angle A E D\)

b the area of \(\triangle A D E\), correct to two decimal places

c \(A D\), correct to two decimal places

d \(\angle A D E\)

e \(\angle A D C\)

f \(\angle D A C\)

\(\mathbf{g}\) the area of \(\triangle A D C\), correct to two decimal places

h the area of the pentagon, correct to two decimal places

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-31(1)}
\end{center}

10 A quadrilateral has diagonals of length \(12 \mathrm{~cm}\) and \(18 \mathrm{~cm}\). If the angle between the diagonals is \(65^{\circ}\), find the area of the quadrilateral, correct to the nearest square centimetre.

11 The sides of a triangle \(A B C\) are enlarged by a factor, \(k\). Use the area formula to show that the area is enlarged by the factor, \(k^{2}\).

12 Prove that the formula Area \(=\frac{1}{2} a b \sin C\) gives the area of a triangle when \(C\) is obtuse.

13 A triangle has sides of length \(8 \mathrm{~cm}, 11 \mathrm{~cm}\) and \(15 \mathrm{~cm}\).

a Find the size of the smallest angle in the triangle, correct to two decimal places.

b Calculate, correct to two decimal places, the area of the triangle.

c Calculate the perimeter of the triangle.

d Let \(s=\) half the perimeter of the triangle. The area of the triangle can be found using Heron's formula: Area \(=\sqrt{s(s-a)(s-b)(s-c)}\), where \(a, b\) and \(c\) are the lengths of the three sides. Use this formula to calculate the area of the triangle, correct to two decimal places.

e Check that your answers to parts \(\mathbf{b}\) and \(\mathbf{d}\) are the same.

\section*{Review exercise}
1 Calculate the value of the pronumeral in each triangle. Give all side lengths correct to two decimal places and all angles correct to one decimal place.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-32}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-32(6)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-32(8)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-32(4)}
\end{center}

2 Find the exact value of the pronumeral in each triangle.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-32(1)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-32(3)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-32(5)}
\end{center}

\(3 A B=8 \mathrm{~cm}, B C=6 \mathrm{~cm}\) and \(A C=12 \mathrm{~cm}\). Find the magnitude of each of the angles of triangle \(A B C\) correct to one decimal place.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-32(7)}
\end{center}

4 A triangular region is enclosed by straight fences of lengths 42.8 metres, 56.6 metres and 72.1 metres.

a Find the angle between the \(42.8 \mathrm{~m}\) and the \(56.6 \mathrm{~m}\) fences, correct to the nearest degree.

b Find the area of the region, correct to the nearest square metre.

5 In a triangle \(A B C, \sin A=\frac{1}{8}, \sin B=\frac{3}{4}\) and \(a=8\). Find, using the sine rule, the value of \(b\).

6 In a triangle, \(A B C, a=5, b=6\) and \(\cos C=\frac{1}{5}\). Find \(c\).

7 Find the area of triangle \(X Y Z\), correct to two decimal places.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-32(2)}
\end{center}

8 For a triangle \(A B C, A C=16.2 \mathrm{~cm}, A B=18.6 \mathrm{~cm}\) and \(\angle A C B=60^{\circ}\). Find, correct to one decimal place:
a \(\angle A B C\)
b \(\angle B A C\)
c the length of \(C B\)
d the area of the triangle

9 The angle of depression from a point \(A\) to a ship at point \(B\) is \(10^{\circ}\). If the distance \(B X\) from \(B\) to the foot of the cliff at \(X\) is \(800 \mathrm{~m}\), find the height of the cliff, correct to the nearest metre.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-33(7)}
\end{center}

10 Calculate the lengths of the unknown sides and the sizes of the unknown angles, correct to two decimal places.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-33(4)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-33(6)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-33(5)}
\end{center}

\(\mathbf{e}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-33(1)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-33(3)}
\end{center}

\(\mathbf{f}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-33}
\end{center}

\section*{Challenge exercise}
1 Write down two formulas for the area of triangle \(A B C\) and deduce the sine rule from those two formulas.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-33(2)}
\end{center}

2 Simi is standing 200 metres due east of Ricardo. From Ricardo, the angle of elevation from the ground to the top of a building due north of Ricardo is \(12^{\circ}\). From Simi, the angle of elevation from the ground to the top of the building is \(9^{\circ}\).

a Let the height of the building be \(h\) metres. Express the following in terms of \(h\) :

i the distance from Ricardo to the foot of the building

ii the distance from Simi to the foot of the building
b Use your answers to part a and Pythagoras' theorem to find the height of the building correct to one decimal place.

c On what bearing is the building from Simi?

3 For triangle \(A B C\), show that

Area \(=\frac{a^{2} \sin B \sin C}{2 \sin A}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-34(1)}
\end{center}

4 Here is an alternative proof of the cosine rule. Assume \(\triangle A B C\) is acute-angled.

a Prove that \(a=b \cos C+c \cos B\).

b Write down corresponding results for \(b\) and \(c\).

c Show that \(a^{2}=a(b \cos C+c \cos B)\) and, using

\includegraphics[max width=\textwidth, center]{2024_02_22_e203939eb913b173edc2g-34(3)}
corresponding results for \(b^{2}\) and \(c^{2}\), prove the cosine rule.

d Check that a similar proof works for an obtuse-angled triangle.

\(5 A B C\) is an isosceles triangle, with \(A B=A C=1\).

Suppose \(\angle B A C=2 \theta\).

a Show that \(B C^{2}=2(1-\cos 2 \theta)\).

b Show that \(B C=2 \sin \theta\).

c Deduce that \(1-\cos 2 \theta=2(\sin \theta)^{2}\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-34}
\end{center}

d Deduce that \(\cos 2 \theta=\cos ^{2} \theta-\sin ^{2} \theta\).

6 a Use the cosine rule to show that

\[
\begin{aligned}
& 1+\cos A=\frac{(b+c)^{2}-a^{2}}{2 b c} \text { and } \\
& 1-\cos A=\frac{a^{2}-(b-c)^{2}}{2 b c}
\end{aligned}
\]

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_e203939eb913b173edc2g-34(2)}
\end{center}

b Let \(s=\frac{a+b+c}{2}\).

Show that \(1+\cos A=\frac{2 s(s-a)}{b c}\)

and \(1-\cos A=\frac{2(s-b)(s-c)}{b c}\)

c Use the fact that \((\sin A)^{2}=1-(\cos A)^{2}\) to show that the square of the area of \(\triangle A B C\) is \(s(s-a)(s-b)(s-c)\) and deduce Heron's formula for the area of a triangle:

\[
\text { Area }=\sqrt{s(s-a)(s-b)(s-c)}
\]

7 Given two sides and a non-included angle, describe the conditions for 0,1 , or 2 triangles to exist.


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