Question

A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30° with the man's eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 60°. What is the distance between the base of the tower and the point P?​

Answer 1

Answer:

your answer and solution is attached!!

Answer 2

Answer:

Distance between the base of the tower and the point P is the  $\frac{3}{2}$ time's distance traveled by the man

Step-by-step explanation:

Recall the values:

tan  30° = $\frac{1}{\sqrt{3} }$

tan 60° = √3

Solution:

Let 'A' be the foot of the tower and 'h' be the height of the tower, then from the diagram we have AB = h.

Let 'x' be the distance between the foot of the tower and the point 'P'

and 'y' be the distance traveled by the man

From the ΔPAB,

tan 30 = $\frac{AB}{AP}$

Substituting the values we get,

$\frac{1}{\sqrt{3} }$ = $\frac{H}{x}$

√3H = x

H = $\frac{x}{\sqrt{3} }$ --------------(1)

Again from ΔQAB,

tan 60 = $\frac{AB}{AQ}$

√3 = $\frac{H}{x-y}$

H = √3(x-y) ---------------(2)

From equations(1) and (2) we get

√3(x-y) = $\frac{x}{\sqrt{3} }$

3(x-y) = x

3x -3y = x

2x = 3y

x = $\frac{3}{2}y$

AP =  $\frac{3}{2}$ X distance traveled by the man

∴ Distance between the base of the tower and the point P is the  $\frac{3}{2}$ times the distance traveled by the man

#SPJ2