Question

The angle of depression of the top and bottom of a building 50 metres high as observed from the top of a tower are 30° and 45° respectively. Find the height of the tower and also the horizontal distance between the building and the tower

Answer 1

Solution :

• The height of the building is 50 m .

• The height of the tower isnt mentioned

• The angles of depression if the top and bottom of the building are 30° and 45° respectively

We have to find the seperation distance between the building and tower horizontally

The simplest way to approach is to take the tan component for both cases .

Suppose the height of the tower is x and the seperation distance is s .

tan 45 = x/s = 1

tan 30 = x-50/s = 1/√3

When x/s is equal to 1 , s = x

Substituting x as s ( because we need to find s)

>> s-50/s = 1/√3

>> s = √3s - 50√3

>> s(1-√3) = -50√3

s = -50√3/(1-√3) = 50√3/(√3-1)

>> 50×1.732/0.732

>> 50 × 2.3661

>> 118.3 m .

Answer : The height of the tower and the horizontal distance between the building and the tower are equal and is 118.3 m

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Answer 2

Given :-

The angle of depression of the top and bottom of a building 50 metres high as observed from the top of a tower are 30° and 45° respectively

To Find :-

Height of the tower and also the horizontal distance between the building and the tower

Solution :-

Let

Building = AB

Tower = CD

Distance between tower and building = AC

AC and BE are parallel. So, AC = BE

tan (30) = DE/BE

1/√3 = DE/BE

√3(DE) = BE

√3DE = BE (..i)

tan 45 = DC/AC

1 = DC/AC

1(AC) = DC

• AC = BE

BE = DC/1 (..ii)

DE√3 = DC/1

DE√3 × 1 = DC

DE × 1.7 = DC

1.7DE = DC

• DC = DE + EC

1.7DE = DE + EC

1.7DE - DE = EC

0.7DE = EC

0.7DE = 50

DE = 50/0.7

DE = 71.4 m

• Height = DC

Height = 71.4 + 50

Height = 121.4 m

Using 1

BE = 71.4 × √3

BE = 71.4√3 m