**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"

**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 = 50√3 ----(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,

tan 60 = ED/AD

√3 = ED/AD

AD = ED/√3 -----(1)

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)

= 4√3

= 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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