Question

The sides of a right angled triangle are in G.P. Find the cosines of the acute angles of the triangle.

Answer 1

$a,b,c$ are in G.P
$b^2=ac$
$c^2=a^2+b^2=a^2+ac$
or ${\left(\frac{a}{c}\right)}^2+\frac{a}{c}-1=0$
$\therefore \frac{a}{c}=\cos B=\frac{-1\pm \sqrt{5}}{2}=\frac{\sqrt{5}-1}{2}$ $\because$ $\cos B$ is +ive
$\therefore \cos A=\sin B=\sqrt{1-{\cos }^{2}B}=\sqrt{(1-\frac{6-2\sqrt{5}}{4})}=\sqrt{\frac{\sqrt{5}-1}{2}}$