JEE Main 2022 (Online) 28th July Evening ShiftQuestion 1
A horizontal park is in the shape of a triangle $OAB$ with $AB=16$. A vertical lamp post $OP$ is erected at the point $O$ such that $\angle PAO=\angle PBO=15^\circ$ and $\angle PCO=45^\circ$, where $C$ is the midpoint of $AB$. Then $(OP)^2$ is equal to
Trigonometric Ratio & Identities
Solution
Concept. Equal angles at $A$ and $B$ give $OA=OB$, so $OC\perp AB$; Pythagoras in $\triangle OCA$ finds $OP$. Formulas used. (1) $\cot15^\circ=2+\sqrt3$, $\cot45^\circ=1$Step 1 - Distances. With $OP=h$, by (1): $OA=OB=h(2+\sqrt3)$ and $OC=h$. Step 2 - Pythagoras. $OA=OB$ and $C$ is the midpoint of $AB$, so $OC\perp AB$ and $AC=8$: $$h^2(7+4\sqrt3)=h^2+64\ \Longrightarrow\ h^2(6+4\sqrt3)=64.$$ Step 3 - Simplify. $$h^2=\frac{32}{3+2\sqrt3}=\frac{32(2\sqrt3-3)}{3}=\frac{32}{\sqrt3}\,(2-\sqrt3).$$ Answer: (B)