Question

On the level ground, the angle of elevation of a tower is 30°. On moving 20 m nearer, the angle of elevation is 60°. The height of the tower is​

Answer 1

$\huge\underline\mathfrak\pink{Answer}$

Let AS be the tower and BD be the ground.

Let h be the height of the tower.

In ΔADB,

$\sf \: tan \: 30 \degree = \frac{h}{20+x} \\ \\ \sf \implies \frac{1}{ \sqrt{3} } = \frac{h}{20+x} \\ \\ \sf \implies \: h= \frac{20+x}{ \sqrt{3} }$

In ΔADB,

$\sf \implies \: tan \: 60 \degree = \frac{h}{x} \\ \\ \sf \implies\: h= \sqrt{3} x \\ \\ \sf \: Now, \sqrt{3} = \frac{20+x}{ \sqrt{3} } \\ \\ \sf \implies \: 3x=20+x \\ \\\sf \implies \: x=10 \\ \\ \sf \implies \: h= \sqrt{3} ×10=10 \sqrt{3} \: m$

Answer 2

Step-by-step explanation:

* this is the language you write first:-let the AB is is the height of the tower and DB is the distance between the foot of the tower and between the observer. suppose observer is at point D after walking 20m . observer is at point C .now, the distance between point C and foot of the tower is x metre.

*Above given is the figure of question

#next step is given in the picture above