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Why the Law of Sines Holds

Part of the Triangles Without Right Angles section of Coddy's Trigonometry journey. Lesson 2 of 37.

Triangle ABC with the height h from C to AB

Draw the height h: a line from C straight down to AB, at a right angle. It splits the triangle into two right triangles. Both right triangles have the same height h. This is why the law works.

In the left triangle, h is opposite A and the hypotenuse is b. So h=b⋅sin⁡(A)h = b\cdot \sin\left(A\right). In the right triangle, h=a⋅sin⁡(B)h = a\cdot \sin\left(B\right). It is the same h, so b⋅sin⁡(A)=a⋅sin⁡(B)b\cdot \sin\left(A\right) = a\cdot \sin\left(B\right). Divide both sides by sin A and by sin B, and you get the law. If B is more than 90°, the same idea works, because B and 180∘−B180^{\circ}-B have the same sine.

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Challenge

In triangle ABC, A = 40°, AC = 12 cm and B = 65°. The height h from C meets AB at D. Find h, then the side x = BC. Done means h and x are in centimetres with 2 decimals.

Then press Check.

Try it yourself

Use the figure to find x,h

Use the figure to find x,h

All lessons in Triangles Without Right Angles