Question

A river flowing East has a
30 ft
tower on one of its banks. On the other side of the river the angle of

elevation from
A
to the top of the tower is
45°
and the angle of elevation from
B
to the top of the tower

is
60°
. Find the width of the river, if
AB ft = 60

is a line on the ground.

Answer 1

This diagram is given as follows;

Height of the tower = CD = 30 ft

Angle of elevation at A = 45°

Angle of elevation at B = 60°

Now,

From ΔBDC,

tan 60° = $\frac{DC}{BC}$

⇒ $\sqrt{3}$ = $\frac{30}{BC}$

⇒ BC = $\frac{30}\sqrt{3} {}$ ft

⇒ BC = $10\sqrt{3}$ ft

Similarly,

From Δ ADC,

tan 45° = $\frac{DC}{AC}$

⇒ 1 = $\frac{30}{AB+BC}$

⇒ AB + $10\sqrt{3}$ = 30

⇒ AB = (30 - $10\sqrt{3}$) ft

∴ width of the river = AB+BC = (30 - $10\sqrt{3} + 10\sqrt{3}$) ft = 30 ft

Ans) The width of the river is 30 ft

A similar answer can be found here :-

https://brainly.in/question/16277999

Answer 2

Given :

Height of the tower = 30 ft

Distance AB = 60 ft

Angle of elevation from A = 45°

Angle of elevation from B = 60°

To find :

The width of the river

Solution :

In ∆OBC,

tan60° = OC / OB

= 30/x

√3 = 30 / x

x = 30/√3 = 30√3 / 3

x = 10√3 ft

Hence, The width of the river is 10√3 ft