Question bank/JEE Main/Mathematics

Let B and C be two points on a circle with center A. Let P be a
point on radius AB such that BC = AP & AB. BP = (AP)^2 and
$\angle CAB = \theta$, then $\theta$ is equal to

A$\frac{\pi}{6}$B$\frac{\pi}{4}$C$\frac{\pi}{5}$D$\frac{\pi}{10}$

Show answer and solution▾

[IMAGE] Let AC = AB = r (radius)
$\cos\theta = \frac{AB^{2} + AC^{2} - BC^{2}}{2AB \cdot AC}$
AB.(AB - AP) = (AP)^2
$AP = \left( \frac{\sqrt{5} - 1}{2} \right)AB$
$cos\theta = \frac{\sqrt{5} + 1}{4}$

Source: Grass-G / M-24 Mathematics Paper 31 Dec.docx