Question

10
Prove that, in a right triangle with a 15° angle, the altitude to the hypotenuse is one fourth of the hypotenuse.

Answer 1

For a better understanding of the solution provided here please go through the diagram in the file attached.

As can be seen in the triangle ABC, the altitude, "a" is AC. The hypotenuse, "h" is BC. The angle $\angle ABC=15^{\circ}$.

Now, we know from the given triangle that:

$Sin(15^{\circ})=\frac{AC}{BC}=\frac{a}{h}$

Now, $Sin(15^{\circ})\approx0.2588\approx0.26\approx0.25$

Therefore, $\frac{a}{h}\approx0.25=\frac{1}{4}$

Therefore we can see that in the given question we can approximately say that the ratio of the altitude to the hypotenuse is 1:4. Equivalently, we can say that the altitide is one fourth of the hypotenuse.

Because, if $\frac{a}{h}=\frac{1}{4}$, then, $a=\frac{1}{4}h$