Question

OBSERVATION A person standing 100 feet from the bottom of a cliff notices a tower on top of the cliff. The angle of elevation to the top of the cliff is 30°, and the angle of elevation to the top of the tower is 58°. How tall is the tower?

Answer 1

Answer:

The tower is 102.26 feet tall.

Step-by-step explanation:

We have drawn the triangle for your reference.

According to the figure point 'A' is the persons eye.

Point 'B' is the bottom of the cliff.

Point 'C' is the top of the cliff.

And Point 'D' is the top of the tower.

Given,

A person standing 100 feet from the bottom of a cliff notices a tower on top of the cliff.

So from diagram we can say that;

Length of AB = 100 ft

The angle of elevation to the top of the cliff is 30°.

So from diagram we can say that;

∠CAB = 30°

the angle of elevation to the top of the tower is 58°.

So from diagram we can say that;

∠DAB = 58°

We have to find the height of the tower i.e. CD.

Solution,

In ΔCAB,

∠CAB = 30°

AB = 100 ft

Now according to trigonometric  ratios;

$tan\theta=\frac{opposite\ side}{adjacent\ side}$

Substituting the values we get;

$tan\ 30\° = \frac{BC}{100}$

Now

$tan\ 30\° = \frac{1}{\sqrt{3}}$

So

$\frac{1}{\sqrt{3}} = \frac{BC}{100}\\\\BC =\frac{100}{\sqrt{3}} = 57.74 \ ft$

In ΔDAB,

∠DAB = 58°

AB = 100 ft

Now according to trigonometric  ratios;

$tan\theta=\frac{opposite\ side}{adjacent\ side}$

Substituting the values we get;

$tan\ 58\° = \frac{BD}{100}$

Now

$tan\ 58\° = 1.60$

So

$1.60 = \frac{BD}{100}\\\\BD =100\times 1.6 = 160 \ ft$

Now BD = BC + CD

CD = BD - BC = 160 - 57.74 = 102.26 ft

Hence The tower is 102.26 feet tall.