Question

If the sides of a right angled triangle are in G.P., then the cosine of the greater acute angle is –

• A. $\frac{2}{1 + \sqrt{5}}$
• B. $\frac{1}{1 - \sqrt{5}}$
• C. $\frac{1 + \sqrt{5}}{2}$
• D. None of these

Answer 1

Answer: (A)

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

Answer 2

Let the sides of the right angled triangle be a, ar, ar² out of which ar² is the hypotenuse, then r > 1.

Now, a² r⁴ = a² + a² r²

or r⁴ – r² – 1 = 0,

\ r² = $\frac{1 \pm \sqrt{5}}{2}$

Q r > 1, \ r² > 1, \ r² = $\frac{1 + \sqrt{5}}{2}$

Angel C is the greater acute angle

Now, cos C = $\frac{a}{ar^{2}} = \frac{1}{r^{2}} = \frac{2}{1 + \sqrt{5}}$.