IB Analysis and Approaches HL · Trigonometry · Section 5 of 5

Sine and cosine rules; area of a triangle

3 worked examples · By Álvaro López

Key idea

When a triangle has no right angle, the ratios of the right-angled triangle are not enough. Two rules, the sine rule and the cosine rule, relate the sides and angles of any triangle, and one formula, $\frac{1}{2} ab sin C$ , gives its area. They appear in Paper 1 with special angles and in Paper 2 in context problems with bearings and heights.

Notation and what "solving a triangle" means

In a triangle $A B C$ the angle $A$ is at vertex $A$ , and the side $a$ is the one opposite it: each side carries the lower-case letter of the opposite vertex. This is what gives meaning to all the formulas that follow.

Triangle ABC with angles A, B and C at the vertices and sides a, b and c opposite them

Solving a triangle means calculating the three sides and the three angles from three pieces of data (with at least one side). Always remember that the angles add up to $18 0^{\circ}$ : if you know two, the third follows directly.

Sine rule

In a triangle, each side is proportional to the sine of the opposite angle. The idea is simple: the longest side is always opposite the largest angle, and the sine captures that proportion.

You only need two of the three ratios each time. It is the right rule when you know a pair "side and opposite angle" and one more piece of that kind: two angles and any side (AAS or ASA), or two sides and an angle opposite one of them (SSA).

A hint of why it works: the height $h$ from $C$ is $h = b sin A$ and also $h = a sin B$ . Equating them, $b sin A = a sin B$ , which is the rule written in another order.

The ambiguous case (SSA)

If you know two sides and an angle that is not between them, the sine rule gives you $sin B$ , and from there you must find $B$ . But one value of $sin B$ is shared by two angles between $0^{\circ}$ and $18 0^{\circ}$ :

$B_{1} = arcsin (sin B) and B_{2} = 18 0^{\circ} - B_{1}$

Sometimes both work and there are two different triangles. The figure shows this: with the angle $A$ , the side $b$ and the side $a$ fixed, the circle of radius $a$ and centre $C$ cuts the base at two points $B_{1}$ and $B_{2}$ .

With angle A, side b and side a, the circle of centre C and radius a cuts the base at B1 and B2, forming two triangles AB1C and AB2C

Cosine rule

When the data do not form a "side–opposite angle" pair (for example two sides and the angle between them, or all three sides), the sine rule does not help. This is where the cosine rule comes in. It generalises Pythagoras' theorem: if the angle is $9 0^{\circ}$ , the term with $cos$ disappears.

The side that stands alone on the left is the one opposite the angle on the right. If $C$ is acute, $cos C > 0$ and $c^{2} < a^{2} + b^{2}$ ; if $C$ is obtuse, $cos C < 0$ and $c^{2} > a^{2} + b^{2}$ .

One advantage of the cosine rule for finding angles: since cosine is one-to-one between $0^{\circ}$ and $18 0^{\circ}$ (a different value for each angle), there is no ambiguous case, and its sign tells you whether the angle is acute or obtuse. That is why, when you know the three sides, calculate the largest angle first with the cosine (the one opposite the largest side) and the others with the sine: you already know they are acute.

Area of a triangle

The height on the side $b$ is $h = a sin C$ , so the area $\frac{1}{2} b h$ becomes the following.

The angle must be between the two sides. If what you have are the three sides, first calculate an angle with the cosine rule, and get its sine with $sin C = 1 - cos^{2} C$ , which is valid because $C$ is between $0^{\circ}$ and $18 0^{\circ}$ and its sine is positive.

Which rule do I use

Look at the three pieces of data you are given, not at the diagram.

Data	Rule	Comment
Two angles and a side	Sine	First the third angle, $18 0^{\circ} - A - B$
Two sides and the angle opposite one of them (SSA)	Sine	Possible ambiguous case
Two sides and the angle between them	Cosine	Gives the third side; then sine or cosine for the angles
The three sides	Cosine	Start with the largest angle

For the area, use any pair of sides with their included angle.

Bearings

In navigation problems, the bearing of one point from another is the angle measured from north, clockwise, and is written with three figures: $04 5^{\circ}$ is north-east, $09 0^{\circ}$ is east, $18 0^{\circ}$ is south and $27 0^{\circ}$ is west.

To convert bearings into angles of the triangle, draw at each point an arrow pointing north (they are all parallel) and use the fact that two parallel lines cut by a transversal form equal or supplementary angles. What you need is the interior angle of the triangle at each vertex.

Remember also the reverse bearing: if $B$ is on bearing $θ$ from $A$ , then $A$ is on bearing $θ + 18 0^{\circ}$ from $B$ (if $θ \geq 18 0^{\circ}$ , subtract $18 0^{\circ}$ ).

A worked example to see it all together. From the port $P$ a boat sails $12$ km on a bearing of $07 0^{\circ}$ to $A$ , and from there $9$ km on a bearing of $13 0^{\circ}$ to $B$ . The reverse bearing from $A$ to $P$ is $07 0^{\circ} + 18 0^{\circ} = 25 0^{\circ}$ , and the interior angle at $A$ is $25 0^{\circ} - 13 0^{\circ} = 12 0^{\circ}$ . Using the cosine rule:

$P B^{2} = 1 2^{2} + 9^{2} - 2 \cdot 12 \cdot 9 cos 12 0^{\circ} = 144 + 81 + 108 = 333 ⟹ P B = 333 \approx 18.25 km$

For the bearing of $B$ from $P$ , the angle at $P$ satisfies $sin P = \frac{9 sin 12 0 ^{\circ}}{333} \approx 0.427$ , so $P \approx 25. 3^{\circ}$ (acute, because $A$ is already obtuse). $B$ lies clockwise from the direction of $A$ , so the bearing is $07 0^{\circ} + 25. 3^{\circ} \approx 09 5^{\circ}$ .

Angles of elevation and depression

The angle of elevation is the angle that the line of sight to an object higher than the observer makes with the horizontal. The angle of depression is the angle that the line of sight to an object lower than the observer makes with the horizontal. Both are measured from the horizontal, not from the vertical, and by parallel lines, the angle of depression from a point equals the angle of elevation from the point below.

Two basic examples. $40$ m from the base of a building, the angle of elevation of the roof is $3 5^{\circ}$ : the height is $40 tan 3 5^{\circ} \approx 28.0$ m. From the top of a $50$ m lighthouse a buoy is seen at an angle of depression of $2 0^{\circ}$ : its horizontal distance from the foot of the lighthouse is $\frac{50}{tan 2 0 ^{\circ}} \approx 137.4$ m.

When the problem has two observations from different points, a non-right triangle is formed: solve it first with the sine (or cosine) rule to find an auxiliary side, and then use the right-angled triangle to get the height.

Worked examples

Example 1 Basic

Solving a triangle with two angles and a side

In the triangle $A B C$ , $A = 3 0^{\circ}$ , $B = 4 5^{\circ}$ and $a = 8$ cm.

a) Find the angle $C$ .

b) Calculate the side $b$ (in exact form).

c) Calculate the area of the triangle (in exact form, in $cm^{2}$ ).

Solution

a) The angles add up to $18 0^{\circ}$ : $C = 18 0^{\circ} - 3 0^{\circ} - 4 5^{\circ} = 10 5^{\circ}$ .

b) The pair $a$ and $A$ is known, and we are asked for $b$ , opposite $B$ . Sine rule:

$\frac{b}{sin 4 5 ^{\circ}} = \frac{8}{sin 3 0 ^{\circ}} ⟹ b = \frac{8 \cdot \frac{2}{2}}{\frac{1}{2}} = 82 cm$

c) The sides $a$ and $b$ form the angle $C$ between them. Use $sin 10 5^{\circ} = sin (6 0^{\circ} + 4 5^{\circ}) = \frac{6 + 2}{4}$ , which comes from the sine of a sum formula:

$Area = \frac{1}{2} ab sin C = \frac{1}{2} \cdot 8 \cdot 82 \cdot \frac{6 + 2}{4} = 82 (6 + 2) = 163 + 16$

The area is $16 + 163 \approx 43.7 cm^{2}$ .

Example 2 Medium

A triangle given by its three sides

A triangle has sides $a = 7$ , $b = 9$ and $c = 12$ .

a) Calculate $cos C$ and decide whether $C$ is acute or obtuse.

b) Calculate the exact value of $sin C$ .

c) Calculate the exact value of the area.

d) Calculate the exact value of $sin A$ .

Solution

a) The angle $C$ is opposite the largest side, $c = 12$ . Cosine rule:

$cos C = \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab} = \frac{49 + 81 - 144}{2 \cdot 7 \cdot 9} = \frac{- 14}{126} = - \frac{1}{9}$

As $cos C < 0$ , the angle $C$ is obtuse ( $C \approx 96. 4^{\circ}$ ).

b) With $sin C > 0$ (because $C$ is between $0^{\circ}$ and $18 0^{\circ}$ ):

$sin C = 1 - \frac{1}{81} = \frac{80}{81} = \frac{4 5}{9}$

c) The sides $a$ and $b$ form the angle $C$ :

$Area = \frac{1}{2} \cdot 7 \cdot 9 \cdot \frac{4 5}{9} = 145 \approx 31.3$

d) Sine rule with the pair $c$ and $C$ :

$\frac{sin A}{7} = \frac{sin C}{12} ⟹ sin A = \frac{7}{12} \cdot \frac{4 5}{9} = \frac{7 5}{27}$

As $a < c$ , the angle $A$ is not the largest, so it is acute (no ambiguity): $A \approx 35. 4^{\circ}$ .

Example 3 Hard

Two sides and an opposite angle: how many triangles?

A triangular plot $A B C$ has $A = 3 0^{\circ}$ , $a = 5$ hm and $b = 8$ hm, where $a$ is the side opposite $A$ and $b$ is the side opposite $B$ .

a) Calculate $sin B$ .

b) Explain why there are two possible triangles and give the two values of $cos B$ .

c) For each triangle, calculate the length of the side $c$ (in exact form).

d) For each triangle, calculate the area (in exact form, in $hm^{2}$ ).

Solution

a) Sine rule:

$sin B = \frac{b sin A}{a} = \frac{8 \cdot \frac{1}{2}}{5} = \frac{4}{5}$

b) Between $0^{\circ}$ and $18 0^{\circ}$ , $sin B = \frac{4}{5}$ is satisfied by two angles: $B_{1} = arcsin \frac{4}{5} \approx 53. 1^{\circ}$ and $B_{2} = 18 0^{\circ} - B_{1} \approx 126. 9^{\circ}$ . Check whether each one fits in a triangle with $A = 3 0^{\circ}$ :

$A + B_{1} \approx 83. 1^{\circ} < 18 0^{\circ}$ : valid.
$A + B_{2} \approx 156. 9^{\circ} < 18 0^{\circ}$ : valid.

Both solutions work. This matches the criterion in the figure: $b sin A = 4 < a = 5 < b = 8$ . As $cos B_{1} > 0$ and $cos B_{2} < 0$ ,

$cos B_{1} = 1 - \frac{16}{25} = \frac{3}{5} cos B_{2} = - \frac{3}{5}$

c) The shortest route is the cosine rule on the angle $A$ (no need for $C$ ):

$a^{2} = b^{2} + c^{2} - 2 b c cos A ⟹ 25 = 64 + c^{2} - 2 \cdot 8 \cdot \frac{3}{2} c$

$c^{2} - 83 c + 39 = 0 ⟹ c = \frac{8 3 \pm 192 - 156}{2} = \frac{8 3 \pm 6}{2} = 43 \pm 3$

Both values are positive: $c_{1} = 43 - 3 \approx 3.93$ hm (the triangle with $B_{2}$ , where $C \approx 23. 1^{\circ}$ ) and $c_{2} = 43 + 3 \approx 9.93$ hm (with $B_{1}$ , $C \approx 96. 9^{\circ}$ ).

That the quadratic equation has two positive roots confirms the ambiguous case: there is one triangle for each root.

d) With the sides $b$ and $c$ and the angle $A$ between them, $Area = \frac{1}{2} b c sin A = \frac{1}{2} \cdot 8 \cdot c \cdot \frac{1}{2} = 2 c$ :

$Area_{1} = 2 (43 - 3) = 83 - 6 \approx 7.86 Area_{2} = 2 (43 + 3) = 83 + 6 \approx 19.86$

In the exam: you earn credit for noticing that $sin B = \frac{4}{5}$ has two solutions and for checking that both fit ( $A + B < 18 0^{\circ}$ ). Giving only the acute angle is the classic mistake. Solving for the side with the cosine rule as a quadratic equation is a good alternative, because it gives both solutions without the risk of forgetting one.

Common mistakes

Summary

Notation: the side $a$ is opposite the angle $A$ , and likewise $b$ and $B$ , $c$ and $C$ . The angles add up to $18 0^{\circ}$ .
Sine rule: $\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$ . Use it with two angles and a side, or with two sides and an angle opposite one of them.
Ambiguous case (SSA): $sin B$ gives two angles, $B_{1}$ and $18 0^{\circ} - B_{1}$ . With $A$ acute there are two triangles if $b sin A < a < b$ ; always check with $A + B < 18 0^{\circ}$ .
Cosine rule: $c^{2} = a^{2} + b^{2} - 2 ab cos C$ , and $cos C = \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab}$ . Use it with two sides and the angle between them, or with the three sides (start with the largest angle).
Area: $\frac{1}{2} ab sin C$ , with $C$ between $a$ and $b$ .
Bearing: angle from north, clockwise, three figures; the reverse bearing adds $18 0^{\circ}$ . Elevation and depression: angles with the horizontal.

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