Question

The sides of a right angled triangle form a geometric progression then cosine of the acute angles will be

• A. $\frac{\sqrt{5}-1}{2}$
• B. $\sqrt{\frac{\sqrt{5}-1}{2}}$
• C. $\frac{2}{\sqrt{5}+1}$
• D. $\sqrt{\frac{\sqrt{5}+1}{2}}$

Answer 1

Answer: (A), (B)

$\sqrt{\frac{\sqrt{5}-1}{2}}$
$\frac{\sqrt{5}-1}{2}$

Let $△ABC$ be right angled at C
$\Rightarrow a^2+b^2=c^2$
also $b^2=ac\left(say\right)$
$\Rightarrow a^2+ac-c^2=0$
$\Rightarrow {\left[\frac{a}{c}\right]}^2+\left[\frac{a}{c}\right]-1=0$
considering $\cos B=t$
$\Rightarrow t^2+t-1=0$
$=t=\left[\frac{\sqrt{5}-1}{2}\right]$ & $\left[\frac{-\sqrt{5}-1}{2}\right]$ (not possible)
$\cos B=\left[\frac{\sqrt{5}-1}{2}\right]$
also $\cos A=\frac{b}{c}=\sqrt{\frac{a}{c}}=\sqrt{\frac{\sqrt{5}-1}{2}}$