GCSE Maths · Pythagoras and Trigonometry · Higher · Lesson 20.8

GCSE Maths: The sine rule (Higher)

Part of the GCSE Maths pythagoras and trigonometry chapter (Higher). Chapter overview · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)

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20.8

The sine rule

Edexcel 1MA1: G22
Grade 6-8

SOH CAH TOA only works in right-angled triangles. The sine rule and the cosine rule work in any triangle. In triangle ABC we use a capital letter for each angle and the matching small letter for the side opposite it: side a is opposite angle A, side b is opposite angle B and side c is opposite angle C.

Each small letter is the side opposite the capital letter of the same name.

The sine rule.

asin A = bsin B = csin C

It says: each side divided by the sine of the angle opposite it gives the same number. You only need two of the three fractions at a time. To find an angle it is easier to turn the rule upside down: sin A ÷ a = sin B ÷ b = sin C ÷ c.

Why it works. Draw the perpendicular height h from C to the side AB. In the right-angled triangle on the left, sin A = h ÷ b, so h = b sin A. In the right-angled triangle on the right, sin B = h ÷ a, so h = a sin B. Both equal h, so b sin A = a sin B. Dividing both sides by sin A sin B gives a ÷ sin A = b ÷ sin B.

The height h = b sin A and also h = a sin B.

When to use the sine rule. You need a matching pair: a side and the angle opposite it, that you know completely. Then:

know two angles and any side (AAS or ASA): find another side. If only two angles are given, find the third angle first (the three angles add to 180°).
know two sides and an angle that is not between them (SSA, where the angle is opposite one of the sides): find another angle.

If you know two sides and the angle between them, or all three sides, there is no matching pair, so you use the cosine rule instead (next section).

The ambiguous case. When you use the sine rule to find an angle, the inverse sine only gives an acute angle. But sin x = sin (180° − x), so the obtuse angle 180° − x has the same sine. Sometimes both angles make a real triangle. For example, with angle A = 30°, side b = 10 cm and side a = 6 cm, the side BC can swing to meet the base line at two different places, B₁ and B₂. You get two different triangles from the same information.

Same angle (30°) and same sides (10 cm and 6 cm), but two possible triangles: AB₁C and AB₂C.

To decide: work out the obtuse possibility 180° − x. If the known angle plus this obtuse angle is less than 180°, a second triangle exists. This only happens when the side opposite the known angle is shorter than the other known side. If the question says the angle is acute (or obtuse), or the diagram shows it, choose that one.

Key ideas

Sine rule: a ÷ sin A = b ÷ sin B = c ÷ sin C. For angles: sin A ÷ a = sin B ÷ b = sin C ÷ c.
Use it when you have a matching pair (side and opposite angle).
Two angles given? Find the third angle first (angles add to 180°).
Finding an angle: the other possible angle is 180° − (calculator answer). Check it fits in the triangle.

Worked example 20.8.1

Work out the length x. Give your answer correct to 1 decimal place.

Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.

Method. The sine rule: each side divided by the sine of its opposite angle gives the same number. Use one complete pair to find the missing side.

Match each side with the angle opposite it. The sine rule links a side with the angle opposite that side. We need one matching pair that we know completely, and one pair with the unknown.
Known pair: side a = 8.5 cm is opposite angle A = 40°.
Wanted pair: side b is opposite angle B = 65°.
Write the sine rule with the unknown on top. Putting the side we want on top means we only need to multiply at the end.
bsin B = asin A
bsin 65° = 8.5sin 40°
Multiply both sides by sin 65°. This gets b on its own.
b = 8.5 × sin 65°sin 40°
Use the calculator (degree mode). Find the two sines, then multiply and divide.
sin 65° = 0.90630… and sin 40° = 0.64278…
b = 8.5 × 0.90630… ÷ 0.64278… = 11.98469…
Round to 1 d.p.. Give the answer to the accuracy asked, with the unit.
b = 12.0 cm

Check. Each side ÷ sine of its opposite angle should be the same number: 8.5 ÷ sin 40° = 13.2 and 12.0 ÷ sin 65° = 13.2. ✓

Watch out. Always pair a side with the angle opposite it (not next to it). Draw arrows from each angle to the side across from it if you are unsure.

Worked example 20.8.2

Work out the length x. Give your answer correct to 1 decimal place.

Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.

Method. The sine rule: each side divided by the sine of its opposite angle gives the same number. Use one complete pair to find the missing side.

Find the third angle. The angles in a triangle add up to 180°. We need angle C, because it is opposite the side we want.
C = 180° − 48° − 71° = 61°
Match each side with the angle opposite it. The sine rule links a side with the angle opposite that side. We need one matching pair that we know completely, and one pair with the unknown.
Known pair: side a = 9.2 cm is opposite angle A = 48°.
Wanted pair: side c is opposite angle C = 61°.
Write the sine rule with the unknown on top. Putting the side we want on top means we only need to multiply at the end.
csin C = asin A
csin 61° = 9.2sin 48°
Multiply both sides by sin 61°. This gets c on its own.
c = 9.2 × sin 61°sin 48°
Use the calculator (degree mode). Find the two sines, then multiply and divide.
sin 61° = 0.87461… and sin 48° = 0.74314…
c = 9.2 × 0.87461… ÷ 0.74314… = 10.82763…
Round to 1 d.p.. Give the answer to the accuracy asked, with the unit.
c = 10.8 cm

Check. Each side ÷ sine of its opposite angle should be the same number: 9.2 ÷ sin 48° = 12 and 10.8 ÷ sin 61° = 12. ✓

Watch out. Always pair a side with the angle opposite it (not next to it). Draw arrows from each angle to the side across from it if you are unsure.

Worked example 20.8.3

In this triangle angle x is acute. Work out the size of angle x, correct to 1 decimal place.

Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.

Method. Use the sine rule with the sines on top: sin(wanted angle) over its opposite side equals sin(known angle) over its opposite side. Then use the inverse sine.

Match each side with the angle opposite it. The sine rule links a side with the angle opposite it. We have one complete pair, and the side opposite the angle we want.
Known pair: side a = 10 cm is opposite angle A = 52°.
Wanted: angle B, which is opposite side b = 7 cm.
Write the sine rule with the sines on top. When we want an angle it is easier to put the sines on top.
sin Bb = sin Aa
sin B7 = sin 52°10
Multiply both sides by 7. This gets sin B on its own.
sin B = 7 × sin 52°10 = 0.55160…
Use the inverse sine to find B. The inverse button turns a sine value back into an angle.
B = sin⁻¹(0.55160…) = 33.47736…°
Look for a second (obtuse) answer. Sine gives the same value for an acute angle and for 180° minus it. The obtuse angle only works if the angles still fit inside 180°.
Second angle = 180° − 33.47736…° = 146.52263…°
Angles so far: 52° + 146.52263…° = 198.52263…°, which is more than 180°, so it is impossible. Only the acute answer works.
Round to 1 d.p. Only one answer is possible.
B = 33.5°

Check. Each sine ÷ its opposite side should match: sin 33.5° ÷ 7 = 0.079 and sin 52° ÷ 10 = 0.079. ✓

Watch out. The inverse sine always gives an acute angle. If the triangle might have an obtuse angle, remember the second answer 180° minus the first.

Worked example 20.8.4

In triangle ABC, angle A = 30°, AC = 10 cm and BC = 6 cm. Work out the possible sizes of angle B, correct to 1 decimal place.

Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.

Method. Use the sine rule with the sines on top: sin(wanted angle) over its opposite side equals sin(known angle) over its opposite side. Then use the inverse sine.

Match each side with the angle opposite it. The sine rule links a side with the angle opposite it. We have one complete pair, and the side opposite the angle we want.
Known pair: side a = 6 cm is opposite angle A = 30°.
Wanted: angle B, which is opposite side b = 10 cm.
Write the sine rule with the sines on top. When we want an angle it is easier to put the sines on top.
sin Bb = sin Aa
sin B10 = sin 30°6
Multiply both sides by 10. This gets sin B on its own.
sin B = 10 × sin 30°6 = 0.83333…
Use the inverse sine to find B. The inverse button turns a sine value back into an angle.
B = sin⁻¹(0.83333…) = 56.44269…°
Look for a second (obtuse) answer. Sine gives the same value for an acute angle and for 180° minus it. A second triangle is possible if the angles still fit inside 180°.
Second angle = 180° − 56.44269…° = 123.55730…°
Angles so far: 30° + 123.55730…° = 153.55730…°, which is less than 180°, so this second triangle can exist.
Round to 1 d.p. and state both answers. Both triangles fit the information given (the ambiguous case).
B = 56.4° or B = 123.6°

Check. Each sine ÷ its opposite side should match: sin 56.4° ÷ 10 = 0.083 and sin 30° ÷ 6 = 0.083. ✓

Watch out. The inverse sine always gives an acute angle. If the triangle might have an obtuse angle, remember the second answer 180° minus the first.

Watch out

Match each side with the angle opposite it. A very common mistake is to use the angle at one end of the side. And when you find an angle, always ask: could the obtuse angle also work? Do not round the sine values in the middle of the calculation: keep them on the calculator.

Exercise 20J The sine rule

Calculator allowed

FWork out the length of the side marked x in each triangle. Give your answers correct to 1 decimal place. All lengths are in centimetres.

(a)

(b)

(c)

(d)

Done

FWork out the length of the side marked x in each triangle. You will need to find a third angle first. Give your answers correct to 1 decimal place. All lengths are in centimetres.

(a)

(b)

(c)

Done

FWork out the size of the angle marked x in each triangle. Each angle x is acute. Give your answers correct to 1 decimal place. All lengths are in centimetres.

(a)

(b)

(c)

Done

PSA and B are two points 200 m apart on a straight shore. A boat is at C. The angle CAB is 52° and the angle CBA is 71°. Work out the distance AC from the boat to A, correct to 1 decimal place.

Done

PSA triangular sail has angles of 55° and 62°. The side opposite the 55° angle is 3.2 m long. Work out the length of the side opposite the 62° angle, correct to 2 decimal places.

Done

PSTwo angles of a triangle are 38° and 79°. The shortest side is 6.1 cm long. Work out the length of the longest side of the triangle, correct to 1 decimal place.

Done

PSB is 500 m due east of A. The bearing of a tower T from A is 035° and the bearing of T from B is 310°. Work out the distance AT, correct to 1 decimal place.

Done

RIn triangle ABC, angle A = 35°, BC = 5 cm and AC = 8 cm. Show that there are two possible triangles by working out both possible sizes of angle B, correct to 1 decimal place.

Done

RRiya wants to find the side marked ? in this triangle using the sine rule. Explain why she cannot start with the sine rule, and say which rule she should use.

Done

RIn triangle ABC the angles at A and B are equal. Use the sine rule to show that a = b.

Done

RChallenge. In triangle ABC, a = 12 cm, angle B = 40° and angle C = 75°. Work out (a) the length of b, (b) the length of c, (c) the perimeter of the triangle. Give each answer correct to 1 decimal place.

(a)b

(b)c

(c)perimeter

Done