Questions on triangles and the sine and cosine rules

Question 1

In triangle ABC, A = 35°, B = 80° and a = 10 cm. Find b to 3 significant figures.

Accepted Answer

The known opposite pair is a and A, so the sine rule gives b = 10 sin 80°/sin 35° = 17.2 cm. M1 for the sine rule, A1 for 17.2 cm. The bigger angle faces the bigger side, and 17.2 > 10 respects that. Each angle pairs with the side opposite it; pairing the 80° with the 10 cm returns 5.82 cm rather than 17.2.

Question 2

A triangle has sides of 8 cm and 11 cm meeting at 30°. Find its exact area.

Accepted Answer

Area = ½ × 8 × 11 × sin 30° = 44 × ½ = 22 cm². M1 for ½ab sin C, A1 for the exact area. Exact here means no rounding is needed at all, since sin 30° = ½.

Question 3

In triangle ABC, a = 5.4 cm, b = 7.9 cm and B = 96°. Find A to 1 decimal place.

Accepted Answer

The known opposite pair is b with B, so the sine rule rearranged for the sine of the unknown angle gives sin A = 5.4 sin 96°/7.9 = 0.6798, and A = 42.8°. M1 for the sine rule with the sines on top, A1 for 42.8°. An inverse sine always leaves a supplement to consider, and here it is 180° − 42.8° = 137.2°. That candidate is not available: the triangle already carries an angle of 96°, and 96° + 137.2° is more than 180°.

Question 4

A triangle has two sides of length 8 cm and 10 cm, and its area is 24 cm². The angle between those two sides is obtuse. Find that angle to 1 decimal place.

Accepted Answer

The area formula uses exactly the two given sides and the angle between them, so 24 = ½ × 8 × 10 × sin C and sin C = 24/40 = 0.6. M1 for rearranging ½ab sin C, A1 for sin C = 0.6. The inverse sine returns 36.9°, and the other angle with that sine is its supplement, 180° − 36.9° = 143.1°, which is the obtuse one the question names. A1 for 143.1°. Both angles give a triangle of area 24 cm² on sides of 8 cm and 10 cm, so the word obtuse is doing the work of selecting one of them; without it the question would have two answers.

Question 5

A triangle has sides of 6 cm and 9 cm meeting at 40°. Find the third side to 3 significant figures.

Accepted Answer

Two sides and the included angle call for the cosine rule. a² = 36 + 81 − 2 × 6 × 9 × cos 40° = 117 − 82.73… = 34.27…, so a = 5.85 cm. M1 for the cosine rule, A1 for 34.27, A1 for 5.85 cm. Hold a² at full accuracy until the final square root; rounding 34.27 to 34 costs the third figure. Note the whole product 2bc cos A is subtracted, rather than cos A on its own.

Question 6

A triangle has sides 5 cm, 6 cm and 7 cm. Find the largest angle to 1 decimal place.

Accepted Answer

The largest angle faces the longest side, so it is the one opposite the 7 cm. The rearranged cosine rule gives cos A = (25 + 36 − 49)/(2 × 5 × 6) = 12/60 = 0.2, so A = 78.5° to 1 decimal place. B1 for choosing the angle opposite the 7 cm, M1 for the rearranged cosine rule, A1 for 78.5°. A positive cosine means an acute angle, so even the biggest angle here is under 90°.

Question 7

In triangle ABC, A = 50°, C = 60° and a = 12 cm. Solve the triangle completely, giving sides to 3 significant figures.

Accepted Answer

First the third angle: B = 180° − 50° − 60° = 70°. Then, using the known opposite pair a and A, b = 12 sin 70°/sin 50° = 14.7 cm and c = 12 sin 60°/sin 50° = 13.6 cm. B1 for B = 70°, M1 for the sine rule, A1 for b = 14.7 cm, M1 for the second application, A1 for c = 13.6 cm. The largest side, b, faces the largest angle, B, as it must. Solve the triangle means every unknown part, so an answer that stops after one side is incomplete.

Question 8

A ship leaves a harbour and sails 12 km on a bearing of 040°. It then changes course and sails 18 km on a bearing of 110°. Find the ship's distance from the harbour, to 3 significant figures.

Accepted Answer

Sketch the two legs. The bearing turns through 110° − 40° = 70°, and the interior angle of the triangle at the point where the ship turns is the rest of the straight line, 180° − 70° = 110°. B1 for the 110° angle. That is two sides with the angle between them, so the cosine rule applies: d² = 12² + 18² − 2 × 12 × 18 × cos 110° = 468 + 147.75 = 615.75, so d = 24.8 km. M1 for the cosine rule, A1 for 615.75, A1 for 24.8 km. Note that cos 110° is negative, so the term that is written as a subtraction adds here, and d² comes out larger than 12² + 18².

Question 9

In triangle ABC, a = 5 cm, b = 9 cm and A = 40°. Show that no triangle with these measurements exists.

Accepted Answer

Apply the sine rule to the opposite pair a with A: sin B = 9 sin 40°/5 = 5.785/5 = 1.157. M1 for the sine rule rearranged for sin B, A1 for 1.157. No angle has a sine greater than 1, so there is no angle B at all and no such triangle exists. B1 for the conclusion with the reason. The picture behind the arithmetic: the perpendicular distance from C to the line AB is 9 sin 40° = 5.79 cm, and a side of 5 cm swung from C is too short to reach the line. At a = 5.79 cm it would just touch, giving one right-angled triangle, and between 5.79 cm and 9 cm two triangles appear.

Question 10

A triangle has sides of 6 cm, 8 cm and 11 cm. Without evaluating any angle, show that the triangle has an obtuse angle, and state which side that angle faces. State also what your argument gives if the longest side is 10 cm instead of 11 cm.

Accepted Answer

Any obtuse angle faces the longest side, so test the angle opposite the 11 cm. The rearranged cosine rule gives cos θ = (6² + 8² − 11²)/(2 × 6 × 8) = (100 − 121)/96. M1 for the rearranged cosine rule applied to the largest angle, A1 for the numerator 100 − 121. Since 6² + 8² = 100 is less than 11² = 121, that numerator is negative, so cos θ < 0 and θ is obtuse, facing the 11 cm side. A1 for the sign argument with the conclusion. With 10 cm in place of 11 cm the numerator is 100 − 100 = 0, so cos θ = 0 and the angle is exactly a right angle: 6, 8, 10 satisfies Pythagoras' theorem, and it is the boundary between the obtuse case and the acute one. B1 for the right angle at 10 cm.

Question 11

From a point A on level ground the angle of elevation of the top T of a vertical tower is 24°. From a point B on the same straight line, 50 m nearer the foot of the tower, the angle of elevation of T is 41°. Find the height of the tower to 3 significant figures.

Accepted Answer

Work in triangle ABT first. Its angle at A is the 24° elevation. The 41° elevation at B is measured on the far side of B, so the angle ABT inside the triangle is 180° − 41° = 139°, and the angle at T is 180° − 24° − 139° = 17°. B1 for angle ABT = 139°, B1 for angle ATB = 17°. The side AB = 50 m faces the angle at T, so the sine rule gives BT = 50 sin 24°/sin 17° = 69.6 m. M1 for the sine rule on that opposite pair, A1 for BT = 69.6 m. The tower is vertical and the ground level, so the height is BT sin 41° = 45.6 m. A1 for 45.6 m.

Question 12

In triangle ABC, a = 7 cm, b = 9 cm and A = 40°. Show that two different triangles fit this data, and find both possible values of B to 1 decimal place.

Accepted Answer

Sine rule: sin B = 9 sin 40°/7 = 0.8264…, so B = 55.7° or its supplement 124.3°. Both must be tested against the angle sum: 40° + 124.3° = 164.3° < 180°, so the obtuse option also leaves room for C. M1 for the sine rule, A1 for sin B = 0.8264, A1 for 55.7°, M1 for taking the supplement, A1 for 124.3°. Two triangles exist.

Question 13

A triangle has sides of 4 cm and 6 cm meeting at 60°. Without a calculator, find the exact length of the third side and the exact area.

Accepted Answer

Cosine rule with cos 60° = ½: a² = 16 + 36 − 48 × ½ = 28, so a = √28 = 2√7 cm. Area = ½ × 4 × 6 × sin 60° = 12 × √3/2 = 6√3 cm². M1 for the cosine rule with cos 60° = ½, A1 for 28, A1 for 2√7 cm, M1 for ½ab sin C, A1 for the exact area.

Question 14

A field is the quadrilateral ABCD, in which AB = 40 m, BC = 55 m, CD = 30 m, DA = 45 m and angle ABC = 68°. Find the length of the diagonal AC, the size of angle ADC, and the area of the field, giving each to 3 significant figures.

Accepted Answer

The diagonal splits the field into two triangles, and triangle ABC has two sides with the angle between them. Cosine rule: AC² = 40² + 55² − 2 × 40 × 55 × cos 68° = 4625 − 1648.3 = 2976.7, so AC = 54.6 m. M1 for the cosine rule, A1 for 54.6 m. Triangle ACD now has three known sides, so the rearranged cosine rule gives cos ADC = (45² + 30² − 2976.7)/(2 × 45 × 30) = −51.73/2700 = −0.01916, and angle ADC = 91.1°. M1 for the rearranged cosine rule, A1 for 91.1°. The field is the two triangles together: ½ × 40 × 55 × sin 68° = 1019.9 m² and ½ × 45 × 30 × sin 91.1° = 674.9 m², so the area is 1690 m² to 3 significant figures. M1 for adding two areas found by ½ab sin C, A1 for 1690 m². Carry AC² = 2976.7 into the second cosine rule rather than the rounded 54.6: the cosine here is −0.019, so small changes decide whether the angle at D lands just past a right angle or just short of it.

Question 15

In triangle ABC the perpendicular from C meets the line AB at N. By finding CN in two ways, prove that a/sin A = b/sin B. Explain why the proof still holds when the angle at A is obtuse.

Accepted Answer

Triangle ANC has its right angle at N and hypotenuse AC = b, with the angle at A between them, so CN = b sin A. Triangle BNC has its right angle at N and hypotenuse BC = a, so CN = a sin B. M1 for CN from one of the two right-angled triangles, A1 for both expressions. The two describe the same perpendicular, so b sin A = a sin B. M1 for equating them. Neither sin A nor sin B is zero in a triangle, since both angles lie strictly between 0° and 180°, so dividing by sin A sin B is legitimate and gives a/sin A = b/sin B. A1 for the division with the printed result. If the angle at A is obtuse, N falls outside the segment AB, beyond A, and the angle of the right-angled triangle at A is then 180° − A rather than A. Its sine is unchanged, because sin (180° − A) = sin A, so CN = b sin A still and the rest of the argument runs as before. B1 for the obtuse case with sin (180° − A) = sin A. The third ratio, c/sin C, comes from dropping the perpendicular from a different vertex.