Question

If the sides of a right-angled triangle are in G.P., then the cosines of the acute angle of the triangle are

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

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{\sqrt{5}-1}{2}$
C $\sqrt{\frac{\sqrt{5}-1}{2}}$

Let the length of the hypotenuse be $h$.

Hence, $h,hsin\theta ,hcos\theta$ are in G.P.

$h^{2}si{n}^2\theta =h^{2}cos\theta$

$si{n}^2\theta =cos\theta$

$1-co{s}^2\theta =cos\theta$

$co{s}^2\theta +cos\theta -1=0$

$cos\theta =\frac{-1\pm \sqrt{5}}{2}$

Since acute angles.

$cos\theta =\frac{\sqrt{5}-1}{2}$

Similarly, $h,hcos\theta ,hsin\theta$ can be in G.P.

Then, $co{s}^2\theta =sin\theta$

$co{s}^2\theta =\sqrt{1-co{s}^2\theta }$

$co{s}^4\theta =1-co{s}^2\theta$

$co{s}^4\theta +co{s}^2\theta -1=0$

Therefore,

$co{s}^2\theta =\frac{-1\pm \sqrt{5}}{2}$

Now, $co{s}^2\theta >0$

$co{s}^2\theta =\frac{\sqrt{5}-1}{2}$

Hence, $cos\theta =\sqrt{\frac{\sqrt{5}-1}{2}}$ ... considering the acute angle.