Question

A TV tower stands vertically on a bank of a canal, with a height of1013 m. From a point on the
other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From
another point on the line joining this point to the foot of the tower, the angle of elevation of the top
of the tower is 30°. Find the distance between the
opposite bank of the canal and the point with 30°
angle of eelevation.

Answer 1

Answer:

Suppose the h is the height of the tower AB and BC = x m

It is given that, width of CD is 20 m,  

According to question,

In triangle \Delta ADB,

\\\Rightarrow \tan 30^o = \frac{AB}{20+x}=\frac{h}{20+x}\\\\\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{20+x}\\\\\Rightarrow 20+x = h\sqrt{3}\\\\\Rightarrow x = h\sqrt{3}-20............(i)

In triangle ACB,

\\\Rightarrow \tan 60^o = \frac{h}{x}=\sqrt{3}\\\\\Rightarrow x= \frac{h}{\sqrt{3}}.............(ii)

On equating eq (i) and (ii) we get:

h\sqrt{3}-20= \frac{h}{\sqrt{3}}

from here we can calculate the value of  h=10\sqrt{3}= 10 (1.732) = 17.32\: m  and the width of the canal is 10 m.

Explanation:

Answer 2

Answer:

ans is 1013 * 2/$\sqrt{3}$

Explanation: