Question

1. A building with height AB and a tower with height CD are shown below. The angle of elevation from point B of the building to the top of the tower is 52°. The angle of elevation from point A of the building to the top of the tower is 25°, The building is 70m shorter than the tower.
a) Find the horizontal distance BD between the structures. [3]
b) Find the distance AC between the structures. [2]
c) Find the height CD of the tower. [4

Answer 1

Answer:

Don't know what you think about it

Answer 2

Given:

Height of building = AB

Height of tower = CD

The angle of elevation from point B of the building to the top of the tower = 52°

The angle of elevation from point A of the building to the top of the tower = 25°

Difference between the height of tower and building = 70 m

To find:

a) The horizontal distance BD between the structures.

b) The distance AC between the structures.

c) The height CD of the tower.

Solution:

(a)

Let the height of tower = h m

Let the distance between the tower and building = x m

In Δ BCD,

tan 52° = $\frac{h}{x}$

As tan 52° is 1.2799, so,

1.2799 = $\frac{h}{x}$

⇒ x = $\frac{h}{1.2799}$                      ...(1)

In Δ ACE,

tan 25° = $\frac{h-70}{x}$

As tan 25° is 0.4663, so,

0.4663 = $\frac{h-70}{x}$

⇒ x = $\frac{h-70}{0.4663}$                    ...(2)

Equating both the equations (1) and (2), we get,

$\frac{h}{1.2799}$ = $\frac{h-70}{0.4663}$

On cross multiplying both the sides,

0.4663 h = 1.2779(h-70)

⇒ 0.4663 h = 1.2779 h - 89.453

⇒ 1.2779 h - 0.4663 h = 89.453

⇒ 0.8116 h = 89.453

⇒ h = $\frac{89.453}{0.8116}$

⇒ h = 110.2180 m

Putting the value of h in equation (1), we get,

x = $\frac{110.2180}{1.2799}$

x = 86.1145 m = BD

Thus, the horizontal distance BD is 86.1145 m.

(b)

sin 25° = $\frac{70}{AC}$

As sin 25° is 0.4226, we get,

AC = $\frac{70}{0.4226}$

AC = 165.6413 m

Thus, the distance AC is 165.6413 m.

(c)

Height of the tower = CD = h = 110.2180 m (as already solved above)

Thus, the height CD of the tower is 110.2180 m.