JEE Main 2021 (Online) 25th July Morning ShiftQuestion 1
A spherical gas balloon of radius $16$ meter subtends an angle $60^{\circ}$ at the eye of the observer $A$ while the angle of elevation of its center from the eye of $A$ is $75^{\circ}$. Then the height (in meter) of the top most point of the balloon from the level of the observer's eye is :
Trigonometric Ratio & Identities
Solution
Concept. The two tangents from the eye make $30^\circ$ each with the line to the centre; the radius is perpendicular to the tangent.Formulas used. (1) $\sin30^\circ=\frac{\text{radius}}{AC}$, $\sin75^\circ=\frac{\sqrt6+\sqrt2}{4}$Step 1 - Distance to the centre. By (1), $AC=\dfrac{16}{\sin30^\circ}=32$. Step 2 - Height of the centre. $32\sin75^\circ=8(\sqrt6+\sqrt2)$. Step 3 - Top point. Add the radius: $8(\sqrt6+\sqrt2)+16=8(\sqrt6+\sqrt2+2)$. Answer: (B)