JEE Main 2016 (Online) 10th AprilQuestion 1
The angle of elevation of the top of a vertical tower from a point A, due east of it is $45^{\circ}$. The angle of elevation of the top of the same tower from a point B, due south of A is $30^{\circ}$. If the distance between A and B is $54\sqrt2$ m, then the height of the tower (in meters), is :
Trigonometric Ratio & Identities
Solution
Concept. Each elevation gives a horizontal distance from the foot $O$; east and south are perpendicular, so the ground triangle $OAB$ has a right angle at $A$. Formulas used. (1) Elevation $\theta$ of a top of height $h$ from distance $d$: $\tan\theta=\dfrac hd$(2) Pythagoras: $OB^2=OA^2+AB^2$ when $\angle OAB=90^\circ$Step 1 - Distances from the foot. By (1): $OA=h\cot45^\circ=h$ and $OB=h\cot30^\circ=h\sqrt3$. Step 2 - The ground triangle. $A$ is due east of $O$ and $B$ is due south of $A$, so $OA\perp AB$:Step 3 - Pythagoras. By (2), $$3h^2=h^2+\left(54\sqrt2\right)^2\ \Longrightarrow\ 2h^2=2\cdot54^2\ \Longrightarrow\ h=54.$$ Answer: (D)