# What will be the Distance of the Object? If

From the top of a tower of height 180 m the angles of depression of two objects on either sides of the tower are 30^{0} and 45^{0} . Then the distance between the objects are

- 180 (3+√3) m
- 180 (3-√3) m
- 180 (√3-1) m
- 180 (√3+1) m

Answer: (3) 180 (√3-1) m

Explanation:-

According to the question,

In the figure-

The height of the tower is BC = 180 m and A & D are two objects which given in the question.

In right angle right triangle BCD,

tan45

^{o}= 180/BD

1 = 180/BD

Then, BD = 180 m ……………………… (1)

Now, In right triangle ABC,

tan30o = 180/(AD + BD)

1/√3 = 180/(AD + 180 ) {from equation (1) }

AD + 180 = √3 x 180 ( by cross multiplication)

AD = 180√3 – 180

AD = 180(√3- 1)

AD = distance between two objects.

Hence, the answer is (3) 180 (√3 – 1) m.