Year 11 Extended · Unit 2 Trigonometric functions and any triangle · Lesson 5 of 6

Two angles and a side fix a triangle. What if you are given two sides and an angle that is not between them?

Split a triangle into two right-angled triangles and a rule for every triangle falls out. Then the lesson asks what two sides and an angle that is not between them can tell you.

Geometry and TrigonometryScientific calculator

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From Lesson 1: sin 140° equals the sine of which acute angle?
    Answer40°. sin (180° − 40°) = sin 40°.
  2. From Year 8: the angles of a triangle are 45° and 80°. What is the third?
    Answer55°. 180 − 125.
  3. From Year 9: in a right-angled triangle the hypotenuse is 20 and an angle is 30°. How long is the side opposite it?
    Answer10. 20 × sin 30°.
Words used in this lesson
sine rule
a side divided by the sine of its opposite angle is the same for all three sides. For example: a ÷ sin A = b ÷ sin B.
opposite
across from: side a is opposite angle A. For example: the side that does not touch the angle.
ambiguous case
a set of measurements that fits two different triangles. For example: two sides and an angle not between them.
perpendicular height
the shortest distance from a corner to the opposite side. For example: the height from C to AB.
obtuse
an angle between 90° and 180°. For example: an angle of 120°.

Smoke on the ridge

Two lookout towers, A and B, stand 5 km apart on a ridge in the Drakensberg. Both see smoke at C. Each tower measures the angle between the line to the other tower and the line to the smoke. There is no right angle anywhere, but you want the distance from each tower to the fire.

Predict first: which measurements do you need, apart from the 5 km? Would the position of the fire be fixed by two angles and a side? And if you knew two sides and an angle that was not between them, could the fire be in two different places?

50
60
8.5height h from C to AB
12.8a ÷ sin A, equal to b ÷ sin B

A = 50°, B = 60°, C = 70°. Height from C: b sin A = 8.5 and a sin B = 8.5. Each side ÷ sine of its opposite angle = 12.8.

Show me the steps
  1. A = 50°, B = 60°, so C = 70°. The height from C is b sin A = 8.5 in the left triangle and a sin B = 8.5 in the right one: the same length found two ways. Each side divided by the sine of its opposite angle is 12.8.
  2. A = 30°, B = 30°: C = 120°, a = b = 6.93. The height is 6.93 × sin 30° = 3.5, and the ratio a ÷ sin A = 6.93 ÷ 0.5 = 13.9. The rule still holds with an obtuse angle at C.
  3. A = 70°, B = 70°: C = 40°. The sides are a = b = 17.54, the height is 17.54 × sin 70° = 16.5, and the ratio is 17.54 ÷ 0.9397 = 18.7. A small angle opposite the fixed side makes everything larger.

The sine rule. In any triangle with sides a, b, c opposite angles A, B, C: a ÷ sin A = b ÷ sin B = c ÷ sin C.

  • Why it works: the height h from C is b sin A in one right-angled triangle and a sin B in the other. So b sin A = a sin B, and dividing both sides by sin A sin B gives a ÷ sin A = b ÷ sin B.
  • Use it when you know a side and the angle opposite it, plus one more side or angle.
  • For an angle, turn the rule over: sin B = b sin A ÷ a. Then B is the angle the calculator gives, or 180° − that angle.
  • An obtuse angle is fine: sin (180° − x) = sin x, as in Lesson 1.

The angles of a triangle add to 180°, so two angles give the third.

8
twotriangles that fit
45.8° or 134.2°angle B

A = 35°, b = 10, a = 8: two triangles fit. B = 45.8° (then C = 99.2°) or B = 134.2° (then C = 10.8°).

Show me the steps
  1. a = 8: the height from C to the base is b sin A = 10 × sin 35° = 5.7, and 8 is more than that but less than b = 10. The arc of length 8 meets the base twice, so two triangles fit: B = 45.8° or 180° − 45.8° = 134.2°.
  2. a = 5: this is less than the height 5.7. The side a cannot reach the base at all, so no triangle fits. In the sine rule, sin B = 10 × sin 35° ÷ 5 = 1.15, which no angle has.
  3. a = 11: this is longer than b = 10. The arc meets the base once on the right side of A. The second angle, 180° − 31.4° = 148.6°, would give 35° + 148.6° = 183.6°, more than 180°, so only B = 31.4° fits.

The sine rule for an angle can give two answers. With the angle A, side a and side b given, B may be the calculator's angle or 180° − that angle. Keep the second one only if it fits: the angles must still add to less than 180°.

Worked examples

In triangle ABC, A = 54°, B = 71° and a = 15. Find b.

  1. Side a is opposite A and side b is opposite B: b ÷ sin 71° = 15 ÷ sin 54°.
  2. b = 15 × sin 71° ÷ sin 54° = 15 × 0.9455 ÷ 0.8090 = 17.5 to one decimal place.

In triangle ABC, A = 40°, a = 7 and b = 9. Find the possible values of B.

  1. sin B = b sin A ÷ a = 9 × sin 40° ÷ 7 = 0.8264. The calculator gives B = 55.7°.
  2. The other angle with that sine is 180° − 55.7° = 124.3°. Test: 40° + 124.3° = 164.3°, less than 180°, so it fits. B = 55.7° or 124.3°, giving C = 84.3° or 15.7°.

Your go. The first step is given. In triangle PQR, P = 48°, p = 15 and q = 11. Find the angle Q.

  1. sin Q = q sin P ÷ p = 11 × sin 48° ÷ 15.
  2. Find Q, then test whether 180° − Q fits with P = 48°.
Check the answer

sin Q = 0.545, so Q = 33.0°. The other angle would be 147.0°, and 48° + 147.0° = 195°, more than 180°. The side p = 15 is longer than q = 11, so only one triangle fits: Q = 33.0°.

With a calculator

Put the calculator in degree mode first and check that sin 30 gives 0.5. For a side, type the top of the fraction and then divide by the sine in the bottom, with brackets, so that the division applies to the whole top: (14 × sin 40) ÷ sin 75.

For an angle, the inverse key (sin⁻¹ or arcsin) gives one angle between 0° and 90° for a positive value. For example sin⁻¹ 0.3 shows 17.5. The second angle, 180 − 17.5 = 162.5, is yours to find, and then to test: does it fit with the angle you were given?

Check before you go on

1. In triangle ABC, A = 50°, a = 12 and b = 9. How many different triangles fit these measurements?

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. In triangle PQR, P = 38°, Q = 65° and p = 14. Find q, to 1 decimal place.

3. In triangle XYZ, x = 15, z = 11 and X = 70°. Find the acute angle Z, to 1 decimal place.

Level 2 · Building

4. In triangle ABC, A = 30°, a = 8 and b = 12. Two triangles fit. Find the larger possible value of B, to 1 decimal place.

5. In triangle ABC, A = 28°, b = 25 cm and the side a is a whole number of centimetres. What is the smallest value of a for which the triangle can be drawn?

Level 3 · Stretch

6. A surveyor on a riverbank in Kerala stands at A and B, 80 m apart, and sights a tree C on the far side. Angle CAB = 62° and angle CBA = 71°. How far is the tree from A, to 1 decimal place?

7. In triangle ABC, A = 115°, B = 25° and a = 30. Find b, to 1 decimal place.

Something to think about

Suppose angle C is 90°. What is the value of c ÷ sin C, and what does that tell you about the common value of the ratio in a right-angled triangle?

One way to think about it

sin 90° = 1, so c ÷ sin C = c, the hypotenuse. In a right-angled triangle the common value of the ratio is the hypotenuse, and in any triangle it is the diameter of the circle that passes through all three corners.

Exit ticket

8. In triangle ABC, b = 9, B = 32° and c = 14. Which statement about angle C is correct?