The Sine and Cosine Rules – OCR FSMQ Additional Maths

1 Labelling and Choosing a Rule

PT2: know the sine and cosine rules and be able to apply them, including the ambiguous case for sine. These rules work in any triangle, not just right-angled ones.

The labelling convention

Capital letters $A$, $B$, $C$ for the angles
Lower-case $a$, $b$, $c$ for the sides opposite them
So side $a$ is always opposite angle $A$.

Which rule? Count what you know.

Sine rule — when you have a complete side–angle pair
Cosine rule — when you do not: two sides and the included angle, or all three sides

You are given	Use	To find
Two angles and any side	Sine rule	Another side
Two sides and a non-included angle	Sine rule	Another angle (check for ambiguity)
Two sides and the included angle	Cosine rule	The third side
All three sides	Cosine rule	Any angle

The angles of a triangle sum to $180°$. If you know two angles, the third is free — often the quickest first step.

2 The Sine Rule

Two forms — use whichever puts the unknown on top

For a side: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$
For an angle: $\dfrac{\sin A}{a} = \dfrac{\sin B}{b}$

Worked Example 1 — Finding a side

In triangle $ABC$, $A = 48°$, $B = 67°$ and $a = 12$ cm. Find $b$ and $c$.

①$C = 180° - 48° - 67° = 65°$

②$\dfrac{b}{\sin 67°} = \dfrac{12}{\sin 48°}$

③$b = \dfrac{12 \times 0.92050}{0.74314} = 14.86$ cm

④$c = \dfrac{12 \sin 65°}{\sin 48°} = \dfrac{12 \times 0.90631}{0.74314} = 14.63$ cm

⑤$b = 14.9$ cm and $c = 14.6$ cm (3 s.f.)

Sense check: the largest side should face the largest angle. Here $B = 67°$ is largest and $b = 14.9$ is the longest side ✓

Worked Example 2 — Finding an angle

In triangle $PQR$, $p = 9$, $q = 7$ and $P = 62°$. Find angle $Q$.

①The unknown is an angle, so put the sines on top:

②$\dfrac{\sin Q}{7} = \dfrac{\sin 62°}{9}$

③$\sin Q = \dfrac{7 \times 0.88295}{9} = 0.68674$

④$Q = 43.4°$ (1 d.p.)

⑤Here $q < p$, so $Q$ must be smaller than $P$ — and it is. The obtuse alternative $136.6°$ is impossible, since $136.6° + 62° > 180°$.

3 The Ambiguous Case

Because $\sin(180° - \theta) = \sin\theta$, an equation such as $\sin Q = 0.8$ has two solutions in a triangle's range: one acute and one obtuse. Sometimes both give a valid triangle. The specification names this explicitly.

When to check for it

Only when the sine rule is used to find an angle.
Then test: does $180° - \theta$, added to the known angle, still leave room for a third angle?

Worked Example 3 — The specification's own example

In triangle $ABC$, $AB = 10$ m, $AC = 8$ m and angle $B = 40°$. Find the two possible values of angle $C$.

①Side $AC = 8$ is opposite $B$; side $AB = 10$ is opposite $C$. So we have a complete pair — sine rule.

②$\dfrac{\sin C}{10} = \dfrac{\sin 40°}{8}$

③$\sin C = \dfrac{10 \times 0.64279}{8} = 0.80349$

④$C = 53.5°$ (the calculator's answer)

⑤Or $C = 180° - 53.5° = 126.5°$

⑥Check both: with $C = 53.5°$, $A = 180 - 40 - 53.5 = 86.5°$ ✓ With $C = 126.5°$, $A = 180 - 40 - 126.5 = 13.5°$ ✓

⑦Both are positive, so both triangles exist: $C = 53.5°$ or $126.5°$ (1 d.p.).

Why two here? The side opposite the known angle ($8$) is shorter than the other given side ($10$). That is the signature of the ambiguous case.

Worked Example 4 — When there is only one triangle

In triangle $ABC$, $a = 11$, $b = 7$ and $A = 58°$. Find angle $B$, and explain why the answer is unique.

①$\sin B = \dfrac{7 \sin 58°}{11} = \dfrac{7 \times 0.84805}{11} = 0.53967$

②$B = 32.7°$ (1 d.p.)

③The alternative is $180° - 32.7° = 147.3°$.

④But $147.3° + 58° = 205.3° > 180°$ — impossible.

⑤So $B = 32.7°$ is the only answer.

The quick test

If the side opposite the known angle is the longer one, the triangle is unique

The cosine rule is never ambiguous. $\cos^{-1}$ returns a unique angle between $0°$ and $180°$, and a negative cosine automatically gives an obtuse angle. If a question could be ambiguous, prefer the cosine rule where you can.

4 The Cosine Rule

Two forms

For a side: $a^2 = b^2 + c^2 - 2bc\cos A$
For an angle: $\cos A = \dfrac{b^2+c^2-a^2}{2bc}$

Note the pattern. The side alone on the left ($a$) is opposite the angle used ($A$). Get that pairing right and the rest follows. If $A = 90°$ then $\cos A = 0$ and the rule collapses to Pythagoras — the cosine rule is Pythagoras with a correction term.

Worked Example 5 — Finding a side

In triangle $ABC$, $b = 13$ cm, $c = 9$ cm and $A = 110°$. Find $a$.

①$a^2 = 13^2 + 9^2 - 2(13)(9)\cos 110°$

②$\cos 110° = -0.34202$ — negative, because the angle is obtuse.

③$a^2 = 169 + 81 - 234 \times (-0.34202) = 250 + 80.03 = 330.03$

④$a = 18.2$ cm (3 s.f.)

Subtracting a negative adds. The obtuse angle makes $a$ longer than Pythagoras would give ($\sqrt{250} = 15.8$), which is exactly right — opening the angle pushes the opposite side further apart.

Worked Example 6 — Finding an angle

A triangle has sides $5$, $8$ and $11$. Find its largest angle.

①The largest angle is opposite the longest side, so set $a = 11$, $b = 5$, $c = 8$.

②$\cos A = \dfrac{25 + 64 - 121}{2(5)(8)} = \dfrac{-32}{80} = -0.4$

③$A = \cos^{-1}(-0.4) = 113.6°$ (1 d.p.)

④The negative cosine correctly told us the angle is obtuse — no ambiguity to resolve.

A useful shortcut: if $a^2 > b^2 + c^2$ the angle $A$ is obtuse; if $a^2 < b^2+c^2$ it is acute. Here $121 > 89$, so obtuse ✓

5 The Area of a Triangle

Area from two sides and the included angle

$\text{Area} = \tfrac12 ab \sin C$
$C$ must be the angle between the sides $a$ and $b$.

Worked Example 7 — Area, then a missing angle

A triangle has sides $9$ cm and $14$ cm with an included angle of $52°$. Find its area. Then find the angle that would give an area of $50$ cm².

①Area $= \tfrac12(9)(14)\sin 52° = 63 \times 0.78801 = 49.6$ cm² (3 s.f.)

②For an area of $50$: $63\sin C = 50$, so $\sin C = 0.79365$.

③$C = 52.5°$ or $C = 180° - 52.5° = 127.5°$

④Both are valid. Two different triangles, one acute-angled at $C$ and one obtuse, have exactly the same area.

This is the ambiguous case appearing in a different guise. Whenever you solve $\sin(\text{something}) = k$ in a geometric context, ask whether the obtuse answer also makes sense.

6 Quick Reference

Labelling

Side $a$ opposite angle $A$.

Sine rule

Needs a complete side–angle pair.

Unknown on top

Choose the form that puts it in the numerator.

Cosine rule

$a^2 = b^2+c^2-2bc\cos A$.

Angle version

$\cos A = \dfrac{b^2+c^2-a^2}{2bc}$.

Ambiguous case

Only when the sine rule finds an angle.

The test

Does $180° - \theta$ leave room for a third angle?

Unique when

The side opposite the known angle is the longer one.

Area

$\tfrac12 ab\sin C$, with $C$ included.

Sense check

Largest angle faces the longest side.

7 Practice Questions

Question 1

In triangle $ABC$, $A = 40°$, $B = 75°$ and $a = 10$. Find $b$.

▶ Show solution

$b = \dfrac{10\sin 75°}{\sin 40°} = \dfrac{10 \times 0.96593}{0.64279}$

$= 15.0$ (3 s.f.)

Question 2

In triangle $ABC$, $b = 6$, $c = 10$ and $A = 35°$. Find $a$.

▶ Show solution

$a^2 = 36 + 100 - 2(6)(10)\cos 35°$

$= 136 - 120 \times 0.81915 = 136 - 98.30 = 37.70$

$a = 6.14$ (3 s.f.)

Question 3

A triangle has sides $4$, $7$ and $9$. Find the angle opposite the side of length $9$.

▶ Show solution

$\cos\theta = \dfrac{16 + 49 - 81}{2(4)(7)} = \dfrac{-16}{56} = -0.28571$

$\theta = 106.6°$ (1 d.p.)

Question 4

Find the area of a triangle with sides $12$ cm and $15$ cm enclosing an angle of $68°$.

▶ Show solution

Area $= \tfrac12(12)(15)\sin 68° = 90 \times 0.92718$

$= 83.4$ cm² (3 s.f.)

Question 5

In triangle $ABC$, $a = 15$, $b = 9$ and $A = 70°$. Find angle $B$ and state whether it is ambiguous.

▶ Show solution

$\sin B = \dfrac{9\sin 70°}{15} = \dfrac{9 \times 0.93969}{15} = 0.56382$

$B = 34.3°$ (1 d.p.)

The alternative $180° - 34.3° = 145.7°$ would give $145.7° + 70° > 180°$ — impossible.

Not ambiguous: the side opposite the known angle ($15$) is the longer one.

Question 6

In triangle $PQR$, $p = 6$, $q = 9$ and $P = 35°$. Find both possible values of $Q$.

▶ Show solution

$\sin Q = \dfrac{9\sin 35°}{6} = \dfrac{9 \times 0.57358}{6} = 0.86037$

$Q = 59.4°$ or $Q = 120.6°$ (1 d.p.)

Check: $59.4 + 35 = 94.4° < 180$ ✓ and $120.6 + 35 = 155.6° < 180$ ✓

Both triangles exist — this is the ambiguous case, as expected since the side opposite the known angle ($6$) is the shorter one.

Question 7

Two sides of a triangle are $8$ cm and $10$ cm, and its area is $30$ cm². Find both possible values of the included angle.

▶ Show solution

$\tfrac12(8)(10)\sin\theta = 30$, so $40\sin\theta = 30$ and $\sin\theta = 0.75$.

$\theta = 48.6°$ or $\theta = 131.4°$ (1 d.p.)

Both are valid angles for a triangle, so both answers must be given.

Question 8

Explain, using the cosine rule, why a triangle with sides $3$, $4$ and $8$ cannot exist.

▶ Show solution

The angle opposite the side of length $8$ would satisfy

$\cos\theta = \dfrac{9 + 16 - 64}{2(3)(4)} = \dfrac{-39}{24} = -1.625$

But $\cos\theta$ can never be less than $-1$, so no such angle exists and the triangle is impossible.

This matches the triangle inequality: $3 + 4 = 7 < 8$, so the two short sides cannot reach across the long one.

Question 9

A ship sails $40$ km on a bearing of $050°$, then $65$ km on a bearing of $140°$. Find its distance from the start.

▶ Show solution

The bearing changes from $050°$ to $140°$, a turn of $90°$ to the right.

So the two legs are at right angles to each other, and the angle at the turning point inside the triangle is $180° - 90° = 90°$.

By Pythagoras (or the cosine rule with $\cos 90° = 0$):

$d^2 = 40^2 + 65^2 = 1600 + 4225 = 5825$

$d = 76.3$ km (3 s.f.)

Question 10

A triangular plot $ABC$ has $AB = 85$ m, $BC = 62$ m and angle $ABC = 73°$.

(a) Find $AC$. (b) Find angle $BAC$. (c) Find the area of the plot. (d) A fence is to run from $B$ perpendicular to $AC$. Find its length, and explain how this gives a second route to the area.

▶ Show solution

(a) Two sides and the included angle, so the cosine rule:

$AC^2 = 85^2 + 62^2 - 2(85)(62)\cos 73°$

$= 7225 + 3844 - 10\,540 \times 0.29237 = 11\,069 - 3081.6 = 7987.4$

$AC = 89.37$ m, so $89.4$ m (3 s.f.)

(b) Now use the sine rule, with $BC = 62$ opposite angle $A$:

$\dfrac{\sin A}{62} = \dfrac{\sin 73°}{89.37}$

$\sin A = \dfrac{62 \times 0.95630}{89.37} = 0.66339$

$A = 41.6°$ (1 d.p.)

No ambiguity: the obtuse alternative $138.4°$ would exceed $180°$ once $73°$ is added. Also, $A$ cannot be the largest angle since $BC$ is not the longest side.

(c) Area $= \tfrac12 \times 85 \times 62 \times \sin 73° = 2635 \times 0.95630$

$= 2520$ m² (3 s.f.)

(d) Let the foot of the perpendicular be $D$. In the right-angled triangle $ABD$:

$\sin(\angle BAC) = \dfrac{BD}{AB}$, so $BD = 85 \sin 41.6° = 85 \times 0.66374 = 56.4$ m.

This length is the perpendicular height of the triangle taken with $AC$ as the base, so

Area $= \tfrac12 \times AC \times BD = \tfrac12 \times 89.37 \times 56.4 = 2520$ m²

This agrees with (c) ✓ — and shows why $\tfrac12 ab\sin C$ works at all: the $a\sin C$ part is the perpendicular height.

The Sine and Cosine Rules (PT2) · OCR FSMQ Additional Maths · Created with MathJax