JEE Advanced 2022 · Paper 2, Mathematics question 9

Angles. $\angle SQR=70^\circ-15^\circ=55^\circ$ and $\angle PRQ=70^\circ-40^\circ=30^\circ$.

• Triangle $PQR$: angles $70^\circ,30^\circ,80^\circ$, so $\alpha=PQ=\dfrac{\sin30^\circ}{\sin80^\circ}=\dfrac1{2\sin80^\circ}$.
• Triangle $QRS$: angles $55^\circ$ (at $Q$), $70^\circ$ (at $R$), $55^\circ$ (at $S$), so $RS=QR=1$.

Triangle $PRS$. By the sine rule, $\dfrac{PS}{\sin40^\circ}=\dfrac{RS}{\sin\theta}$, so
$$\beta\sin\theta=RS\sin40^\circ=\sin40^\circ.$$

Value.
$$4\alpha\beta\sin\theta=\frac{2\sin40^\circ}{\sin80^\circ}=\frac{2\sin40^\circ}{2\sin40^\circ\cos40^\circ}=\sec40^\circ\approx1.305.$$
This lies in $(0,\sqrt2)$ and in $(1,2)$, but not in $(\sqrt2,3)$ or $(2\sqrt2,3\sqrt2)$. (A), (B).