D. Trigonometric Functions · Trigonometric Ratios, Identities and Laws

The Sine Law in Oblique Triangles

D1.6About 70 minutes

Learning goals

I can explain why the sine law also works when the triangle has an obtuse angle.
I can use the sine law to find sides and angles in any oblique triangle.
I can decide whether an unknown angle found with sin⁻¹ should be acute or obtuse.
I can solve problems modelled by oblique triangles with the sine law.

Words to know

oblique triangle
A triangle with no right angle. It can be acute (all angles less than 90°) or obtuse (one angle greater than 90°).

obtuse angle
An angle greater than 90° and less than 180°.

sine law
In any △ABC, a/sin A = b/sin B = c/sin C, or sin A/a = sin B/b = sin C/c.

solve a triangle
Find all the unknown sides and angles.

Remember from Grade 10

In Grade 10 you developed the sine law for acute triangles by drawing an altitude and writing its length two ways:

a/sin A = b/sin B = c/sin C

You used it when you knew a side and its opposite angle (a complete pair) plus one more side or angle. Every triangle then was acute. Now that you can find sines of angles bigger than 90° (lesson 3), you can use the sine law in obtuse triangles too.

Why it still works when an angle is obtuse

Suppose ∠A is obtuse. The altitude from C now falls outside the triangle, onto the extension of side BA at a point D.

In right △ACD, the angle at A is 180° − A, so h = b sin(180° − A).
From lesson 3, sin(180° − A) = sin A (an angle and its supplement have the same sine). So h = b sin A.
In right △BCD, h = a sin B.

So b sin A = a sin B, which rearranges to a/sin A = b/sin B, exactly as before.

When ∠A is obtuse, the altitude h from C lands outside the triangle. Because sin(180° − A) = sin A, the sine law comes out the same.

Worked example

Example 1: Two angles and a side

In △ABC, ∠A = 112°, ∠B = 31° and c = 15 cm. Solve the triangle.

∠C = 180° − 112° − 31° = 37°. Now c and ∠C are a complete pair.
a/sin 112° = 15/sin 37°, so a = 15 sin 112° ÷ sin 37° ≈ 23.1 cm.
b/sin 31° = 15/sin 37°, so b = 15 sin 31° ÷ sin 37° ≈ 12.8 cm.
Check: the largest angle (112°) is opposite the longest side (a). ✓

∠C = 37°, a ≈ 23.1 cm, b ≈ 12.8 cm

Example 1 drawn to scale. Side a, opposite the obtuse angle, is the longest side.

Finding an angle: acute or obtuse?

When you use the sine law to find an angle, sin⁻¹ always gives an acute angle. But you know that sin β = sin(180° − β), so the true angle might be the obtuse one. How do you decide?

A triangle can have at most one obtuse angle. If the triangle already has an obtuse angle, every other angle is acute.
The smallest angle is opposite the shortest side, and the smaller of two angles is opposite the shorter side. An angle opposite a side that is shorter than another known side must be acute (a triangle can't have two angles of 90° or more).
If a diagram or the context says the angle is obtuse, use 180° − β.

If none of these settle it, there may be two possible triangles. That is the ambiguous case, the topic of lesson 10.

Worked example

Example 2: Finding an angle in an obtuse triangle

In △ABC, ∠C = 125°, c = 20 cm and a = 12 cm. Solve the triangle.

Complete pair: c and ∠C. sin A/12 = sin 125°/20, so sin A = 12 sin 125° ÷ 20 ≈ 0.4915.
∠C is already obtuse, so ∠A is acute: ∠A ≈ 29.4°.
∠B ≈ 180° − 125° − 29.4° = 25.6°.
b = 20 sin 25.6° ÷ sin 125° ≈ 10.5 cm. (Use the unrounded angle in your calculator.)

∠A ≈ 29.4°, ∠B ≈ 25.6°, b ≈ 10.5 cm

Worked example

Example 3: When the diagram says obtuse

In △ABC, ∠A = 20°, a = 9 cm and b = 20 cm. A diagram shows that ∠B is obtuse. Find ∠B, ∠C and c.

sin B = 20 sin 20° ÷ 9 ≈ 0.7600.
sin⁻¹(0.7600) ≈ 49.5°. But ∠B is obtuse, so ∠B ≈ 180° − 49.5° = 130.5°.
∠C ≈ 180° − 20° − 130.5° = 29.5°.
c = 9 sin 29.5° ÷ sin 20° ≈ 12.9 cm.

∠B ≈ 130.5°, ∠C ≈ 29.5°, c ≈ 12.9 cm. Without the diagram, ∠B = 49.5° would also have been possible. Lesson 10 shows how to spot this.

Worked example

Example 4: An island on a lake

Points A and B are 600 m apart on a straight shoreline road. An island C is in the lake. Surveyors measure ∠CAB = 28° and ∠CBA = 130°. How far is the island from B? What is the shortest distance from the island to the road (the line through A and B)?

∠C = 180° − 28° − 130° = 22°. Side AB = 600 m is opposite ∠C.
BC is opposite ∠A: BC = 600 sin 28° ÷ sin 22° ≈ 751.9 m.
AC is opposite ∠B: AC = 600 sin 130° ÷ sin 22° ≈ 1227.0 m.
The shortest distance from C to line AB is the altitude from C. In the right triangle with hypotenuse AC and angle 28° at A: altitude = AC sin 28° ≈ 576.0 m.

The island is about 752 m from B, and about 576 m from the road.

Just for fun

The sine law works in every triangle, acute or obtuse. It is one of the few laws that never asks what angle you're coming from.

Hands-on activity

Test the sine law on an obtuse triangle

You need: a ruler, a protractor, a sharp pencil, a calculator and a sheet of paper.

Draw an obtuse triangle with one angle between 110° and 140° and sides between 6 cm and 18 cm. Label it A, B, C with the obtuse angle at A.
Measure all three sides (to the nearest millimetre) and all three angles (to the nearest degree).
Calculate a ÷ sin A, b ÷ sin B and c ÷ sin C.
Extend side BA past A with a light line. Draw the altitude from C to that extension and measure it (h). Check that h ≈ b sin A ≈ a sin B.

What to look for: the three ratios in step 3 should agree within a few percent, even though one angle is obtuse. The altitude lies outside the triangle, just like in the diagram above.

Your turn

In △ABC, ∠A = 118°, ∠B = 27° and a = 20 cm. Find b.

In △ABC, ∠B = 105°, ∠C = 40° and b = 14 m. Find c.

In △ABC, ∠C = 130°, c = 25 cm and b = 16 cm. Find ∠B.

In △ABC, ∠A = 35°, ∠C = 25° and b = 30 cm. Find ∠B, a and c.

Explain why a triangle can have at most one obtuse angle.

Use a calculator to compare sin 130° and sin 50°. Why does this matter for the sine law?

More practice: warm-up

sin 150° equals the sine of which acute angle?

Two angles of a triangle are 110° and 40°. Find the third.

A triangle has angles 100°, 50° and 30°. Which side is longest?

Solve a/sin 120° = 10/sin 30° exactly.

sin B = 0.5 and ∠B is obtuse. Find ∠B.

More practice: on your own

Solve △ABC: ∠A = 25°, ∠B = 125°, a = 8 cm.

In △PQR, ∠P = 140°, p = 30 cm and q = 12 cm. Find ∠Q, ∠R and r.

In △ABC, ∠A is obtuse, a = 15 cm, b = 9 cm and ∠B = 22°. Find ∠A and ∠C.

Points A and B are 2.0 km apart on a straight shore. From A, an island is at 35° to AB; from B it is at 118° to BA. How far is the island from B?

Leo used the sine law and got sin A = 0.656, so he wrote ∠A = 41°. The diagram shows ∠A is obtuse. Correct his answer.

Challenge

∠C is obtuse in △ABC. Draw the altitude from A to the extension of BC. Use it to show that b/sin B = c/sin C.

Two fire towers A and B are 15 km apart. A fire F is spotted with ∠FAB = 21° and ∠FBA = 140°. How far is the fire from each tower, and how far is it from the straight line through the towers?

A triangular sail has angles of 105°, 45° and 30°. Its longest edge is 6.0 m. Find its perimeter.

Check your understanding

Quiz

Why does the sine law still work when ∠A is obtuse?

In △ABC, ∠A = 112°, ∠B = 31° and c = 15 cm. What is a?

In △ABC, ∠C = 125°. You find sin A = 0.4915. What is ∠A?

sin⁻¹ gives 49.5°, but the diagram shows the angle is obtuse. The angle is:

In any triangle, the longest side is opposite:

Which information lets you start with the sine law right away?

For parents and teachers

Big idea (D1.6): the Grade 10 sine law works in obtuse triangles too, because sin(180° − A) = sin A. The new challenge is that sin⁻¹ only gives an acute angle, so students must decide whether the angle they want is acute or obtuse.

How to help: Ask for a rough sketch before any calculation, and ask "Which is the largest angle? Is it opposite the longest side?" as a final check.

Common difficulties:

Accepting the calculator's acute angle when the angle is clearly obtuse.
Rounding an angle early and then using the rounded value to find a side. Keep full values in the calculator.
Pairing a side with an angle that is not opposite it.