# PROBLEMS INVOLVING ANGLE OF ELEVATION AND DEPRESSION

Problem 1 :

From the top of a tree of height 13 m the angle of elevation and depression of the top and bottom of another tree are 45° and 30° respectively. Find the height of the second tree. (3 = 1.732)

Solution :

AB = 13 m  =  DC

In triangle AED,

In triangle ABC,

tan 30  =  AB/BC

1/3  =  13/BC

BC  =  133  ----(2)

By applying the value of AD in (1), we get

ED  =  13

height of second tree  =  ED + DC

=  133 + 13

=  13(3 + 1)

=  13(1.732 + 1)

=  13(2.732)

=  35.516 m

Hence the height of the second tree is 35.52 m

Problem 2 :

A man is standing on the deck of a ship, which is 40 m above water level. He observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30° . Calculate the distance of the hill from the ship and the height of the hill. (√3 = 1.732)

Solution :

In triangle AED,

In triangle ABC,

tan 30  =  AB/BC

1/√3  =  40/BC

BC  =  40√3  -----(2)

(1)  =  (2)

40√3  =  ED/√3

ED  =  40√3(√3)

=  40(3)

ED  =  120 m

Height of hill  =  ED + DC

=  120 + 40

=  160 m

Distance from hill to ship  =  40√3

=  40(1.732)

=  69.28 m

Problem 3 :

If the angle of elevation of a cloud from a point ‘h’ meters above a lake is θ1 and the angle of depression of its reflection in the lake is θ2 . Prove that the height that the cloud is located from the ground is h(tan θ1+ tan θ2)/(tan θ2 - tan θ1)

Solution :

In triangle AED,

AD  =  x cot θ -----(1)

tan θ2  =  (2h + x)/AD

AD  =  (2h + x)/tan θ2

AD  =  (2h + x)cot θ2 -----(2)

(1) = (2)

x cot θ1  =  (2h + x)cot θ2

x cot θ1  =  2h cot θ2 + x cot θ2

x (cot θ1 - cot θ2) =  2h cot θ2

x (tan θ2 - tan θ1)/(tan θ1 tan θ2) =  2h (1/tan θ2)

2h  =  [xtan θ2(tan θ2 - tan θ1)]/(tan θ1 tan θ2)

2h  =  x(tan θ2 - tan θ1)/tan θ1

x  =  2h tan θ1/(tan θ2 - tan θ1)

height of cloud  =  x + h

=  [2h tan θ1/(tan θ2 - tan θ1)] + h

=  [2h tan θ1+ h(tan θ2 - tan θ1)]/(tan θ2 - tan θ1)]

=  h[tan θ1tan θ2]/(tan θ2 - tan θ1)]

Hence proved.

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