Question

If the sides of a right-angled triangle are in AP then what are the sines of the acute angles?

Answer 1

Let $\angle C={90}^{\circ }$ being greatest and $B={90}^{\circ }-A$.

The sides are a - d, a and a + d.

$(a+d{)}^2=(a-d{)}^2+a^2$ (using Pythagoras theorem)



$\therefore 4ad-a^2=0\Rightarrow a=4d$

Hence the sides are 3d, 4d, 5d.

$sinA=\frac{BC}{AB}=\frac{a-d}{a+d}=\frac{3d}{5d}=\frac{3}{5}$

$sinB=\frac{AC}{AB}=\frac{a}{a+d}=\frac{4d}{5d}=\frac{4}{5}$