Question

form a point on the ground 60m away from the foot of the tower the angle of elevation of the top of the tower is 30° the angle of elevation of the top of water tank( on top of the tower) is 45° find height of the tower and depth of tack

Answer 1

This is my solution.....You may note that I have rationalised (h) in the above solution...the answer can come without rationalising also...!

Thank You !

Answer 2

Answer: The height of the tower is 20√3=34.64 m and the depth of tack is 60-34.64=25.35 m.

Step-by-step explanation:

Since we have given that

Angle of elevation of the top of the tower = 30°

Angle of elevation of the top of water tank = 45°

Distance from a point on the ground away from the foot of the tower = 60 m

So, ABD is a right triangle.

BC is the height of tower and AB is the depth of the tack.

CD = 62 m

As shown in the figure :

Consider ΔBCD,

$\tan 30^\circ=\frac{BC}{CD}\\\\\frac{1}{\sqrt{3}}=\frac{BC}{60}\\\\BC=\frac{60}{\sqrt{3}}\\\\BC=\frac{60}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\\\\BC=\frac{60\sqrt{3}}{3}\\\\BC=20\sqrt{3}\ m$

Consider ΔACD,

$\tan 45^\circ=\frac{AC}{CD}\\\\1=\frac{AC}{60}\\\\AC=60\ m$

Hence, the height of the tower is 20√3=34.64 m and the depth of tack is 60-34.64=25.35 m.