Fundamentals of Trigonometry

Lessons

Senior 4 · Mathematics

Fundamentals of Trigonometry

5 lessons

Lesson 5 of 5

Solving triangles: the sine and cosine rules, and applications

25 min

Triangles without a right angle

A pilot who changes course, a surveyor who measures a hillside, or an engineer who anchors a pole on a slope meets triangles with no right angle, where SOH-CAH-TOA does not apply directly. Two rules, both consequences of the unit-circle definitions, solve every such triangle: the cosine rule and the sine rule. In a triangle $A B C$ we write $a, b, c$ for the sides opposite the angles $A, B, C$ .

Important

The cosine rule. In any triangle $A B C$ ,

a^2 = b^2 + c^2 - 2bc\cos A, \qquad b^2 = a^2 + c^2 - 2ac\cos B, \qquad c^2 = a^2 + b^2 - 2ab\cos C .

Proof. Place $A$ at the origin and $B$ at $(c, 0)$ . By the unit-circle definitions, $C$ is at $(b\cos A, b\sin A)$ , whether $A$ is acute or obtuse. Then

a^2 = BC^2 = (b\cos A - c)^2 + (b\sin A)^2 = b^2(\cos^2 A + \sin^2 A) - 2bc\cos A + c^2 = b^2 + c^2 - 2bc\cos A . \;\blacksquare

When $A = 90^\circ$ , $\cos A = 0$ and the rule becomes Pythagoras' theorem.

Important

The sine rule and the area formula. In any triangle $A B C$ ,

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}, \qquad \text{Area} = \frac{1}{2}bc\sin A = \frac{1}{2}ac\sin B = \frac{1}{2}ab\sin C .

Proof. In the figure, the height from $C$ is $h = b\sin A$ (the $y$ -coordinate of $C$ ). Looking from $B$ , the same height is $h = a\sin B$ . So $b\sin A = a\sin B$ , that is $\frac{a}{\sin A} = \frac{b}{\sin B}$ ; using another height gives the third ratio. And the area is $\frac{1}{2} \times \text{base } c \times h = \frac{1}{2}bc\sin A$ . $\blacksquare$

Which rule?

You know	Use first
two sides and the angle between them (SAS)	cosine rule, for the third side
three sides (SSS)	cosine rule, rearranged: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$
two angles and a side (ASA or AAS)	angle sum $180^\circ$ , then the sine rule
two sides and an angle not between them (SSA)	sine rule, and check for a second triangle

Worked example 1: SAS and SSS

(a) In triangle $A B C$ , $b = 7$ , $c = 5$ and $A = 120^\circ$ . Find $a$ , $B$ , $C$ and the area. (b) A triangle has sides 6, 8 and 11. Find its largest angle.

Solution

$a^2 = 49 + 25 - 70\cos 120^\circ = 74 + 35 = 109$ , so $a \approx 10.44$ . Then $\sin B = \frac{7\sin 120^\circ}{10.44} \approx 0.5806$ , $B \approx 35.5^\circ$ (acute, since $A$ is already obtuse), and $C \approx 180^\circ - 120^\circ - 35.5^\circ = 24.5^\circ$ . Area $= \frac{1}{2}(7)(5)\sin 120^\circ \approx 15.16$ .
The largest angle faces the longest side: $\cos C = \frac{6^2 + 8^2 - 11^2}{2(6)(8)} = \frac{-21}{96} \approx -0.2188$ , so $C \approx 102.6^\circ$ . The negative cosine tells you at once that the angle is obtuse.

Worked example 2: AAS, and the ambiguous case

(a) $A = 40^\circ$ , $B = 65^\circ$ , $a = 12$ . Solve the triangle. (b) $a = 6$ , $b = 8$ , $A = 40^\circ$ . Find $B$ .

Solution

$C = 75^\circ$ . $b = \frac{12\sin 65^\circ}{\sin 40^\circ} \approx 16.92$ and $c = \frac{12\sin 75^\circ}{\sin 40^\circ} \approx 18.03$ .
$\sin B = \frac{8\sin 40^\circ}{6} \approx 0.8571$ . There are two angles with this sine: $B \approx 59.0^\circ$ or $B \approx 180^\circ - 59.0^\circ = 121.0^\circ$ . Both leave room for $C$ ( $40^\circ + 121.0^\circ \lt 180^\circ$ ), so two triangles fit the data: $C \approx 81.0^\circ$ or $C \approx 19.0^\circ$ .

Definition

Bearing: the direction of one point from another, given as an angle measured clockwise from north, written with three figures: east is $090^\circ$ , south $180^\circ$ , and south-west $225^\circ$ .

Angle of elevation / depression: the angle between the horizontal and the line of sight to an object above / below the observer (as in S3).

Inclined plane: a surface at an angle $\alpha$ to the horizontal. A slope of $p %$ rises $p$ units for every 100 units horizontally, so $\tan\alpha = \frac{p}{100}$ .

Worked example 3: air navigation

A light aircraft flies from airstrip $P$ on a bearing of $060^\circ$ for 300 km to $Q$ , then on a bearing of $170^\circ$ for 400 km to $R$ . How far is $R$ from $P$ , and on what bearing must the aircraft fly to return directly to $P$ ?

Solution The bearing of $P$ from $Q$ is $060^\circ + 180^\circ = 240^\circ$ , so $\angle PQR = 240^\circ - 170^\circ = 70^\circ$ . Cosine rule:

PR^2 = 300^2 + 400^2 - 2(300)(400)\cos 70^\circ \approx 167\,915, \qquad PR \approx 409.8\ \text{km}.

Sine rule: $\sin\angle QPR = \frac{400\sin 70^\circ}{409.8} \approx 0.9173$ , so $\angle QPR \approx 66.5^\circ$ (acute, because $\angle QPR + \angle PRQ = 110^\circ$ ). The bearing of $R$ from $P$ is $060^\circ + 66.5^\circ = 126.5^\circ$ , so the return bearing from $R$ to $P$ is $126.5^\circ + 180^\circ \approx 306.5^\circ$ .

Worked example 4: two angles of elevation

From a point $A$ on level ground, the angle of elevation of the top $T$ of a tower is $30^\circ$ . From $B$ , 50 m nearer the tower, it is $45^\circ$ . Find the height of the tower.

Solution In triangle $A B T$ : $\angle ABT = 180^\circ - 45^\circ = 135^\circ$ , so $\angle ATB = 180^\circ - 30^\circ - 135^\circ = 15^\circ$ . Sine rule: $BT = \frac{50\sin 30^\circ}{\sin 15^\circ} \approx 96.59\,\mathrm{m}$ . Then in the right triangle $B F T$ , $h = BT\sin 45^\circ \approx 68.3\,\mathrm{m}$ .

Worked example 5: a pole on an inclined plane

A vertical pole 8 m tall stands on a road with a 10% slope. A wire from the top of the pole is fixed to the road 15 m from the foot of the pole, measured straight down the slope. How long is the wire?

Solution The slope makes an angle $\alpha = \tan^{-1} 0.1 \approx 5.71^\circ$ with the horizontal. The pole is vertical, so the angle between the pole and the downhill direction of the road is $90^\circ + 5.71^\circ = 95.71^\circ$ . Cosine rule:

w^2 = 8^2 + 15^2 - 2(8)(15)\cos 95.71^\circ \approx 289 + 23.9 = 312.9, \qquad w \approx 17.7\ \mathrm{m}.

Warning

Three traps. (1) In the SSA case, $\sin^{-1}$ on a calculator gives only the acute angle; always test $180^\circ - B$ as well. (2) Find the largest angle with the cosine rule, not the sine rule, because only the cosine shows whether it is obtuse. (3) Bearings are clockwise from north; mathematical angles are anticlockwise from the $x$ -axis.

Tip

Always draw and label a diagram first, and mark the angle you know inside the triangle. In bearing problems, draw a north line at every point; the angle between two legs of a journey is then a difference or a sum of bearings.

Radford Mathematics: Bearings Trigonometry Problem with Cosine Rule and Sine Rule. Watch for how the angle inside the triangle is found from the two bearings before either rule is used.

✦

Practice

Give lengths to 3 significant figures and angles to 1 decimal place.

In triangle $A B C$ , $a = 9$ , $b = 12$ and $C = 50^\circ$ . Find $c$ and the area.
A triangle has sides 5, 7 and 10. Find its largest angle.
In triangle $P Q R$ , $P = 48^\circ$ , $Q = 72^\circ$ and $r = 15$ . Find $p$ and $q$ .
In triangle $A B C$ , $a = 10$ , $b = 7$ and $A = 50^\circ$ . Show that only one triangle is possible, and find $B$ .
A ship sails 20 km on a bearing of $040^\circ$ , then 35 km on a bearing of $115^\circ$ . Find its distance from the start.
From the top of a cliff 80 m high, the angles of depression of two boats in line with the foot of the cliff are $25^\circ$ and $40^\circ$ . How far apart are the boats?
A ramp rises at a 12% slope. Find its angle with the horizontal, and the height gained along 25 m of ramp.
Prove that the area of a triangle is $\frac{1}{2}ab\sin C$ , and use it to find the area of an equilateral triangle of side 6.

Show answersHide answers

$c^2 = 81 + 144 - 216\cos 50^\circ \approx 86.16$ , $c \approx 9.28$ ; area $= \frac{1}{2}(9)(12)\sin 50^\circ \approx 41.4$ .
$\cos C = \frac{25 + 49 - 100}{70} \approx -0.3714$ , $C \approx 111.8^\circ$ .
$R = 60^\circ$ . $p = \frac{15\sin 48^\circ}{\sin 60^\circ} \approx 12.9$ , $q = \frac{15\sin 72^\circ}{\sin 60^\circ} \approx 16.5$ .
$\sin B = \frac{7\sin 50^\circ}{10} \approx 0.5362$ , so $B \approx 32.4^\circ$ or $147.6^\circ$ ; but $50^\circ + 147.6^\circ \gt 180^\circ$ , so only $B \approx 32.4^\circ$ .
The angle between the legs is $180^\circ - (115^\circ - 40^\circ) = 105^\circ$ . $d^2 = 400 + 1225 - 1400\cos 105^\circ \approx 1987.3$ , $d \approx 44.6$ km.
Distances from the foot: $\frac{80}{\tan 25^\circ} \approx 171.6$ m and $\frac{80}{\tan 40^\circ} \approx 95.3$ m, so about 76.2 m apart. (Or with the sine rule in the triangle at the top.)
$\alpha = \tan^{-1} 0.12 \approx 6.8^\circ$ ; height $= 25\sin 6.84^\circ \approx 2.98$ m.
Height to side $a$ is $b\sin C$ , so area $= \frac{1}{2}a \cdot b\sin C$ . Equilateral: $\frac{1}{2}(6)(6)\sin 60^\circ = 9\sqrt{3} \approx 15.6$ .

Quick check

You know two sides of a triangle and the angle between them. Which rule finds the third side directly?

Key points

Cosine rule: $a^2 = b^2 + c^2 - 2bc\cos A$ (SAS, SSS).
Sine rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ (AAS, ASA, SSA with care).
Area $= \frac{1}{2}ab\sin C$ .
Bearings are clockwise from north; a $p %$ slope has $\tan\alpha = \frac{p}{100}$ .

1Triangles without a right angle

2Which rule?

Triangles without a right angle

Which rule?