Question

The angle of elevation of the top of a tower from a certain point is 30°. if the observer moves 40 m towards the tower, the angle of elevation of the top of the tower increases by 15°. the height of the tower is

Answer 1

your solution

Hope it help you

Answer 2

Let AB be the tower of height h metres, C and D be the two points of observation.Then,

$\sf{CD=20m}$

$\sf{\angle{ACB}=30°}$

$\sf{\angle{ADB}=30°+15°=45°}$

From right ∆ABD,we have

$\sf{\tan45°= \frac{AB}{BD} = 1 = \frac{h}{BD = BD}= h}$

From right ∆ABC, we have

$\longrightarrow$$\sf{\tan30° = \frac{AB}{BC} = \frac{h}{20 + BD} }$

$\longrightarrow$$\sf{ \frac{1}{ \sqrt{3} } = \frac{h}{20 + h} }$

$\longrightarrow$$\sf{\sqrt{3h } = 20 + h}$

$\longrightarrow$$\sf{( \sqrt{3} - 1)h = 20}$

$\therefore$ $\sf{h=\frac{20}{√3-1}×\frac{√3+1}{√3+1}}$

$\longrightarrow$$\sf{10(√3+1)m}$

Hence, the height of the tower is 10(√3+1)m.