JEE Main 2020 (Online) 6th September Morning ShiftQuestion 1
The angle of elevation of the top of a hill from a point on the horizontal plane passing through the foot of the hill is found to be $45^{\circ}$. After walking a distance of $80$ meters towards the top, up a slope inclined at an angle of $30^{\circ}$ to the horizontal plane, the angle of elevation of the top of the hill becomes $75^{\circ}$. Then the height of the hill (in meters) is $\_\_\_\_\_\_\_\_$ .
Trigonometric Ratio & Identities
Answer$80$
Solution
Concept. Start at the origin with the top at $(h,h)$; after $80$ m up a $30^\circ$ slope the observer is at $(40\sqrt3,40)$. Formulas used. (1) $\tan75^\circ=2+\sqrt3$Step 1 - New elevation. By (1), $$\frac{h-40}{h-40\sqrt3}=2+\sqrt3.$$ Step 2 - Solve. $h-40=(2+\sqrt3)h-80\sqrt3-120$, so $h(1+\sqrt3)=80(1+\sqrt3)$ and $h=80$. Answer: 80