JEE Main 2013 (Offline)Question 1
$ABCD$ is a trapezium such that $AB$ and $CD$ are parallel and $BC\perp CD$. If $\angle ADB=\theta$, $BC=p$ and $CD=q$, then $AB$ is equal to:
Trigonometric Ratio & Identities
Solution
Concept. Work in triangle $ABD$: we know $BD$, the angle $\theta$ at $D$, and the angle at $B$ (alternate angles). Formulas used. (1) Sine rule: $\dfrac{AB}{\sin\angle ADB}=\dfrac{BD}{\sin\angle DAB}$(2) $\sin(\theta+\varphi)=\sin\theta\cos\varphi+\cos\theta\sin\varphi$Step 1 - The picture. $BD=\sqrt{p^2+q^2}$. Let $\angle BDC=\varphi$, so $\sin\varphi=\frac{p}{\sqrt{p^2+q^2}}$, $\cos\varphi=\frac{q}{\sqrt{p^2+q^2}}$. Since $AB\parallel DC$, $\angle ABD=\varphi$.Step 2 - Angle at $A$. $\angle DAB=\pi-\theta-\varphi$, so $\sin\angle DAB=\sin(\theta+\varphi)$. By (2), $$\sin(\theta+\varphi)=\frac{q\sin\theta+p\cos\theta}{\sqrt{p^2+q^2}}.$$ Step 3 - Sine rule. By (1), $$AB=\frac{BD\sin\theta}{\sin(\theta+\varphi)}=\frac{\left(p^2+q^2\right)\sin\theta}{p\cos\theta+q\sin\theta}.$$ Answer: (A)