The angle of elevation of the top of a tower from a certain point is $30^\circ$. If the observer moves 20 m towards the tower, the angle of elevation of the top of the tower increases by $15^\circ$. The height of the tower is

### Concept & Difference of Distances When an observer moves towards an object, the difference between the initial and final distances from the base equals the distance walked. $$ \text{Distance Walked} = \frac{h}{\tan(\theta_{\text{initial}})} - \frac{h}{\tan(\theta_{\text{final}})} $$ ### Step-by-Step Solution 1. Given: Initial angle = $30^\circ$. Distance moved towards tower = $20$ m. Angle increases by $15^\circ$, making the new angle $30^\circ + 15^\circ = 45^\circ$. 2. Let the height of the tower be $h$. 3. Initial distance from tower ($d_1$): $$ d_1 = \frac{h}{\tan(30^\circ)} = h\sqrt{3} $$ 4. Final distance from tower ($d_2$): $$ d_2 = \frac{h}{\tan(45^\circ)} = h(1) = h $$ 5. The distance moved is the difference between these two distances: $$ d_1 - d_2 = 20 $$ $$ h\sqrt{3} - h = 20 $$ 6. Factor out $h$ and solve: $$ h(\sqrt{3} - 1) = 20 $$ $$ h = \frac{20}{\sqrt{3} - 1} $$ 7. Rationalize the denominator by multiplying top and bottom by $(\sqrt{3} + 1)$: $$ h = \frac{20(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2} = \frac{20(\sqrt{3} + 1)}{3 - 1} = \frac{20(\sqrt{3} + 1)}{2} = 10(\sqrt{3} + 1) $$ 8. Substitute $\sqrt{3} \approx 1.732$: $$ h = 10(1.732 + 1) = 10(2.732) = 27.32 \text{ m} $$ 9. The closest given option is $27.3$ m. ### Exam Strategy & Shortcut For a $30^\circ \to 45^\circ$ shift, memorize the formula: $h = \frac{d}{2}(\sqrt{3} + 1)$, where $d$ is the distance walked. Plugging in $d = 20$, you instantly get $10(\sqrt{3} + 1)$, bypassing the algebraic derivation entirely. ### Common Pitfall Misinterpreting "increases by $15^\circ$" as meaning the new angle is exactly $15^\circ$. Always add the increase to the original angle ($30 + 15 = 45^\circ$). ### Final Answer Therefore, the correct answer is **27.3 m**.