Question

Topic :- Applications of Trigonometry

Very simple Qns ..

Try to explain abt the diagram very clearly ..

The horizontal distance between two towers is 30meters from foot of first tower the angle of elevation of top of second tower is 60°. From top of second tower angle of depression of top of first is 30°. Find the height of small tower.


Kind request :- Explain the diagram clearly​

Answer 1

Answer:

20√3 m

Step-by-step explanation:

Let,

AC be the second tower.

ED be the first tower.

Given, CD = 30m

→ BE = CD = 30m

Let, ED = x

→ ED = BC = x

AB = y

$\sf\angle ABE = 30^{\circ}$

$\sf\angle ADC = 60^{\circ}$

In ∆ ABE :-

tan30° = AB/BE

1/√3 = y/30

y = 30/√3

y = 10√3m

In ∆ ADC :-

tan60° = x + y/30

√3 = x + 10√3/30

x + 10√3 = 30√3

x = 20√3 m

→ Height of the small tower = ED = x = 20√3 m.

Used Formula :-

1) Opposite interior angles are equal

2) tan30° = 1/√3

3) tan60° = √3

Note :- Refer to above attachment.

Answer 2

Step-by-step explanation:

Topic = Applications of Trigonometry

QUESTION :

• The horizontal distance between two towers is 30meters from foot of first tower the angle of elevation of top of second tower is 60°. From top of second tower angle of depression of top of first is 30°. Find the height of small tower.

given :

• horizontal distance between two towers = 30 m

• elevation of top of second tower = 60 °

• tower angle of depression of top of first is = 30°

to find :

• Find the height of small tower= ?

• Find the height of small tower = ?

solution :

please check the attached file where you find the diagram

• tan 60° = h/ 30

• tan 30° = h-h /30

• 30° = h/30 - h/30

• h/30 = 60° - 30°

• h/30 = 3 -1 / √3

• horizontal = 30× 2 /√3

• horizontal = 20√3m

thus, horizontal distance between two towers =20√3m