# ANGLE OF ELEVATION AND DEPRESSION WORKSHEET

Angle of Elevation and Depression Worksheet :

Here we are going to see, some example problems based on angle of elevation and depression.

To find questions 1 to 3, please visit the page "Problems involving angle of elevation and depression"

## Angle of Elevation and Depression Worksheet - Questions

Question 4 :

The angle of elevation of the top of a cell phone tower from the foot of a high apartment is 60° and the angle of depression of the foot of the tower from the top of the apartment is 30° . If the height of the apartment is 50 m, find the height of the cell phone tower. According to radiations control norms, the minimum height of a cell phone tower should be 120 m. State if the height of the above mentioned cell phone tower meets the radiation norms.

Solution : In triangle BEC,

tan θ  =  Opposite side / Adjacent side

tan 60  =  EC/BC

3  =  (50 + x)/BC

BC  =  (50 + x)/3  ----(1)

In triangle ABD,

tan 30  =  AB / BC

1/√3   =  50/BC

BC  =  503  ----(2)

(1)  =  (2)

(50 + x)/3  =  50√3

50 + x  =  50√3√3

50 + x  =  150

x  =  150 - 50

x  =  100

Height of cell phone tower = 100 + 50

=  150 m

Hence the cell phone tower meets the radiation norms.

Question 5 :

The angles of elevation and depression of the top and bottom of a lamp post from the top of a 66 m high apartment are 60° and 30° respectively. Find (i) The height of the lamp post. (ii) The difference between height of the lamp post and the apartment. (iii) The distance between the lamp post and the apartment. (3 = 1.732)

Solution : In triangle AED,

In triangle ABC,

tan 30  =  AB/BC

1/√3  = 66/BC

BC  =  66√3  -----(2)

(1)  =  (2)

ED/√3  =  66√3

ED  =  66(3)

ED  =  198

(i)  Height of lamp post  =  ED + DC

=  198 + 66

=  264 m

(ii) The difference between height of the lamp post and the apartment

=  264 - 66

=  198 m

(iii) The distance between the lamp post and the apartment

BC  =  66√3

=  66(1.732)

=  114.31 m

Question 6 :

Three villagers A, B and C can see each other across a valley. The horizontal distance between A and B is 8 km and the horizontal distance between B and C is 12 km. The angle of depression of B from A is 20° and the angle of elevation of C from B is 30° . Calculate : (i) the vertical height between A and B. (ii) the vertical height between B and C. (tan 20° = 0.3640, 3 = 1.732)] Solution : In the left side triangle,

tan 20  =  Vertical distance of AB/8

0.3640  =  Vertical distance of AB/8

Vertical distance of AB  =  0.3640(8)

Vertical distance of AB  =  2.91 km

In the right side triangle,

tan 30  =  Vertical distance of BC/12

1/3  =  Vertical distance of BC/12

Vertical distance of BC  =  12/√3

=  12(3/3)

=  43

=  4(1.732)

=  6.928

Vertical distance of BC  =  6.93 km After having gone through the stuff given above, we hope that the students would have understood how to do problems on angle of elevation and depression.

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