JEE Main 2022 (Online) 25th July Morning ShiftQuestion 1
A tower $PQ$ stands on a horizontal ground with base $Q$ on the ground. The point $R$ divides the tower in two parts such that $QR=15$ m. If from a point $A$ on the ground the angle of elevation of $R$ is $60^\circ$ and the part $PR$ of the tower subtends an angle of $15^\circ$ at $A$, then the height of the tower is :
Trigonometric Ratio & Identities
Solution
Concept. $R$ is seen at $60^\circ$ and $P$ at $60^\circ+15^\circ=75^\circ$ from the same point $A$. Formulas used. (1) $\tan60^\circ=\sqrt3$, $\tan75^\circ=2+\sqrt3$Step 1 - $AQ$. $\tan60^\circ=\dfrac{15}{AQ}$, so $AQ=5\sqrt3$. Step 2 - Height. $\angle PAQ=75^\circ$, so $$PQ=5\sqrt3\tan75^\circ=5\sqrt3(2+\sqrt3)=10\sqrt3+15=5(2\sqrt3+3).$$ Answer: (A)