From the top of a 15 m high building, the angle of elevation of the top of a tower is found to be 30° . From the…
CBSE Class 10 Maths PYQ · Applications of Trig · Double Triangle · 5 Marks · March 2024 · Standard
Solve it yourself first — then press ↓ or tap Show Solution. Use ←→ for previous / next question.
515 Marks · March 2024 · Standard
From the top of a $15$ m high building, the angle of elevation of the top of a tower is found to be $30^\circ$. From the bottom of the same building, the angle of elevation of the top of the tower is found to be $60^\circ$. Find the height of the tower and the distance between tower and the building.
Show SolutionHide Solution↓
Correct figure Let CD be the building and AB be the tower. $\tan 30^\circ = \frac{h}{DE} = \frac{h}{x} = \frac{1}{\sqrt{3}}$ ----- (i) $\Rightarrow x = h\sqrt{3}$ $\tan 60^\circ = \frac{h+15}{x} = \sqrt{3}$ ----- (ii) $\Rightarrow h + 15 = x\sqrt{3}$ Solving (i) and (ii) to get $x = 7.5\sqrt{3}$ m or $\frac{15\sqrt{3}}{2}$ m and $h = \frac{15}{2} = 7.5$ m Hence height of the tower is $7.5 + 15 = 22.5$ m
