Question

11. A TV tower stands vertically on a bank of a canal. 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 20 m away from this point on the line joing this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see Fig. 9.12). Find the height of the tower and the width of the canal.
​​

Answer 1

This is the correct answer .....I was solving but time!!!

Answer 2

Answer:

$\bf\fbox\purple{Height of the tower i.e AB is 10√3}$

Step-by-step explanation:

Solution.

Let CB be x

Therefore, DB = 20 + x

In ∆ABD

$\tan( {30}^{0} ) = \frac{AB}{BD} \\ \\ \frac{1}{ \sqrt{3} } = \frac{AB}{20 + x} \\ \\ \frac{20 + x}{ \sqrt{3} } = AB \\ \\ \frac{(20 + x) \sqrt{3} }{ \sqrt{3} \times \sqrt{3} } = AB \\ \\ \frac{20 \sqrt{3} + x \sqrt{3} }{3} = AB - - - - 1$

____________________________

In ∆ ABC

$\tan( {60}^{0} ) = \frac{AB}{BC} \\ \\ \sqrt{3} = \frac{AB}{x} \\ \\ \sqrt{3} = \frac{20 \sqrt{3} + x \sqrt{3} }{3 \times x} (from \: 1) \\ \\ 2x = 20 \\ \\ x = 10$

WIDTH OF THE CANAL = CB = 10 m

Height of the tower = AB

$AB = \frac{20 + x}{ \sqrt{3} } \\ \\ \frac{(20 + 10) \sqrt{3} }{ \sqrt{3} \times \sqrt{3} } \\ \\ 10 \sqrt{3} \: answer$

$\huge\underline\mathcal\red{ 10√3 Answer}$

${\huge{\fcolorbox{blue}{pink}{Galaxy Fanfan}}}$