## Height and Distance

#### Height and Distance

1. The angle of elevation of the top of a hill from each of the vertices A, B, C of a horizontal triangle is α. The height of the hill is :

1. As per the given details of above question , we have
The distance of the foot from each vertex = h cotα.
∴ The foot is at the circumcentre of the triangle.
∴ R = h cotα

##### Correct Option: B

As per the given details of above question , we have
The distance of the foot from each vertex = h cotα.
∴ The foot is at the circumcentre of the triangle.
∴ R = h cotα

 ∴ R = h tanα = a tanα = a tanα . cosecα . 2sinα 2

1. The angle of elevation of the top of an unfinished tower at a point distant 120 m from its base is 45°. If the elevation of the top at the same point is to be 60°, the tower must be raised to a height :

1. Let us draw the figure from the given question.
Given :- OB = h + x

 h + x = tan 60° = √3 120

 h + x = √3 (120) .

##### Correct Option: B

Let us draw the figure from the given question.
Given :- OB = h + x
In figure , we have

 h + x = tan 60° = √3 120

 h + x = √3 (120) ...........( 1 )

 Also, h = tan 45° = 1 . 120

 ∴ h = 120 m

From eq. ( 1 ) ,
∴ h + x = 120 + x = 120 √3
 ∴ x = 120 ( √3 - 1 ) m .

1. A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is 60°. When he retires 40 m from the bank, he finds the angle to be 30°. The breadth of the river is :

1. Let us draw the figure from the given question.
Let, OA denote the breadth of the river.
Given :- AB = 40 m

##### Correct Option: C

Let us draw the figure from the given question.
Let, OA denote the breadth of the river.
Given :- AB = 40 m
In triangle OAP , we have

 OP = tan60° = √3 OA

 ∴ OP = √3 OA.

 Also, In triangle OBP , we have OP = tan 30° = 1 OA + 40 √3

 ∴ OA + 40 = √3 OP = √3 ( √3 OA ) = 3 OA.

 ⇒ 2OA = 40 ⇒ OA = 20 m.

Hence , the breadth of the river = 20 m .

1. The angle of elevation of the top of a tower at a point G on the ground is 30°. On walking 20 m towards the tower, the angle of elevation becomes 60°. The height of the tower is equal to :

1. Let us draw the figure from the given question.
Let, AB = h be the height of the tower. Let, GA = x.

##### Correct Option: D

Let us draw the figure from the given question.
Let, AB = h be the height of the tower. Let, GA = x and GH = 20 m.
In triangle BGA ,

 Then, h = tan30° = 1 . x √3

 ∴ h = x ......... ( 1 ) √3

 In triangle BHA , Also, h = tan60° = √3 . x - 20

 ∴ h = √3 ( x - 20 )

 ∴ x = √3 ( x - 20 ) [ using ( 1 ) ] √3

 ⇒ x = 3 ( x - 20 ) = 3x - 60

 ⇒ 2x = 60 ⇒ x = 30 .
Putting the value of x in eq. ( 1 ) , we get
 ∴ h = 30 = 10 √3 m . √3

1. The angle of elevation of the top of a tower from a point 20 m away from its base is 45°. The height of the tower is :

1. Let us draw the figure from the given question.
Let us assume the angle of elevation ∠MŌP = 45°.
Given :- MO = 20 m
In triangle MOP ,

 tan45° = PM MO

 Clearly, h = tan45 = 1 20

##### Correct Option: B

From the given figure , we can see that
Let us assume the angle of elevation ∠MŌP = 45°. and The height of the tower = h
Given :- MO = 20 m
In triangle MOP ,

 tan45° = PM MO

 ⇒ h = tan45° = 1 20

∴ h = 20 m