Heights And Distances

The standard heights-and-distances approach is to model the scene with two right triangles that share the unknown vertical height, converting each angle of elevation into a tangent relation between height and horizontal distance. Let the tower height be $h$ and let the foot of the nearer observation point lie at horizontal distance $x$ from the base. From the nearer point the elevation is $60^{\circ }$, so $\tan 60^{\circ }=\frac{h}{x}$ gives $x=\frac{h}{\sqrt{3}}$. From the farther point, which is $40$ metres more distant, the elevation is $30^{\circ }$, so $\tan 30^{\circ }=\frac{h}{x+40}$, leading to $x+40=h\sqrt{3}$. Eliminating $x$ by substituting the first relation gives $\frac{h}{\sqrt{3}}+40=h\sqrt{3}$, hence $40=h\sqrt{3}-\frac{h}{\sqrt{3}}=\frac{3h-h}{\sqrt{3}}=\frac{2h}{\sqrt{3}}$. Solving this single equation yields $h=20\sqrt{3}$ metres. Option $40\sqrt{3}$ doubles the answer by dropping the factor of two during elimination. Option $20$ ignores the $\sqrt{3}$ scaling that the tangents introduce. Option $60$ results from misreading the angle pairing. This follows the classic two-observation tower pattern of JEE Advanced. As a final consistency check, $h\approx 34.6$ metres produces a near distance of $20$ metres and a far distance of $60$ metres, whose difference is exactly the given $40$ metre gap.