Finding an Angle

Part of the Triangles Without Right Angles section of Coddy's Trigonometry journey. Lesson 5 of 37.

Here we know two sides and an angle that is not between them: A = 30°, a = 8, b = 6. The law gives $\sin\left(B\right) = \frac{3}{8}$ . A calculator says B is 22.02°. But 157.98° has exactly the same sine.

$Triangle ABC with A = 30°, BC = 8 and AC = 6$

B and $180^{\circ}-B$ always have the same sine. The three angles must add up to 180°. So keep an angle only if the two angles you have add up to less than 180°. Here 30° + 157.98° is more than 180°, so B is about 22.02°. If the law gives a sine above 1, there is no triangle.

Challenge

In triangle ABC, A = 50°, a = BC = 12 cm and b = AC = 9 cm. Find the angle θ at B. Keep only angles that make a triangle. Done means θ is in degrees with 2 decimals.

Then press Check.

Try it yourself

$Use the figure to find theta$