Question

Let in a right angled triangle, the smallest angle be \(\theta\). If a triangle formed by taking the reciprocal of its sides is also a right angled triangle, then \(\sin\theta\) is equal to :

• A. \(\frac{\sqrt{5}+1}{4}\)
• B. \(\frac{\sqrt{5}-1}{2}\)
• C. \(\frac{\sqrt{2}-1}{2}\)
• D. \(\frac{\sqrt{5}-1}{4}\)

Hint:

Take the hypotenuse as 1, so the sides are \(\sin\theta\), \(\cos\theta\), 1. The reciprocal triangle has hypotenuse \(1/\sin\theta\) (the largest reciprocal). Apply Pythagoras to it.

Answer 1

The Correct Option is B

Let the hypotenuse be 1 and the smallest side \(a = \sin\theta\), the other leg \(b = \cos\theta\), so \(a^2 + b^2 = 1\) and \(a \lt b \lt 1\).
Reciprocal sides: \(\frac{1}{a} \gt \frac{1}{b} \gt 1\). The largest, \(\frac{1}{a}\), must be the hypotenuse of the new right triangle:
\(\frac{1}{a^2} = \frac{1}{b^2} + 1\).
Put \(u = a^2\), \(b^2 = 1 - u\): \(\frac{1}{u} = \frac{1}{1-u} + 1 = \frac{2-u}{1-u}\).
\(1 - u = 2u - u^2 \Rightarrow u^2 - 3u + 1 = 0 \Rightarrow u = \frac{3-\sqrt{5}}{2}\) (the root less than 1).
Since \(\left(\frac{\sqrt{5}-1}{2}\right)^2 = \frac{6-2\sqrt{5}}{4} = \frac{3-\sqrt{5}}{2}\), we get \(\sin\theta = \frac{\sqrt{5}-1}{2}\).
Answer: (2) \(\frac{\sqrt{5}-1}{2}\)