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