Question

Prove that, in a right triangle with a 15° angle, the altitude to the hypotenuse is one fourth of the hypotenuse.

Answer 1

Consider the attached figure. If AB has length 1, then BC has length sin(15°) and CD (the altitude of triangle ABC) has length sin(15°)·cos(15°).

By the double angle formula for sin(α), ...

... sin(2α) = 2sin(α)cos(α)

Rearranging, this gives

... sin(α)·cos(α) = sin(2α)/2

We have

... CD = sin(15°)·cos(15°) = sin(2·15°)/2

... CD = sin(30°)/2 = (1/2)/2 = 1/4

That is, the altitude, CD, is 1/4 the hypotenuse, AB, of triangle ABC.