Question

The angle of elevation of the top of a tower from a point 100m away from the base of the tower is 45°. The angle of elevation of the top of the flag staff mounted on the top of the tower is 60°. The height of the flag staff is.......... (Use √3 =1.73)

Answer 1

Refer to the attachment

Let AB is the tower of height h metre and AC is flagstaff of height x metre.

So, ∠APB = 45° and ∠BPC = 60°

tan60° = x + h/100

√3 = x + h/100

x = 100√3 - h

Now,

tan45° = h/100

1 = h/100

h = 100.

Substituting the values of h in x., we get,

x = 100 √3 - 100

x = 100(√3 - 1)

x = 100(1.73 - 1)

x = 100 × 0.73

x = 73.

Therefore, the height of the flagstaff = 73 metres

Answer 2

Solution

Given that :-

• The angle of elevation of the top of a tower from a point 100m away from the base of the tower is 45
• The angle of elevation of the top of the flag staff mounted on the top of the tower is 60

Find :-

• The Height of the flag

Explanation

Let, here

• Flag(AB) = x meter
• Tower ( BC) = H meter
• Base(CD) = 100 m
• <BDC = 45°
• <ADC = 60°

First take , Right ∆ BDC

==> tan 45° = perpendicular/Base = BC/CD

==> tan 45° = H/100

[ tan 45° = 1 ]

==> 1 = H/100

==> H = 100 m ----------(1)

Again, take Right ∆ ADC

==> tan 60° = AC/CD = ( AB + BC)/CD

[ tan 60° = √3 ]

==> √3 = (X+H)/100

Keep Value of H

==> √3 = (X + 100)/100

==> 1.73 × 100 = X + 100

==> X + 100 = 173

==> X = 173 - 100

==> X = 73

Hence

• Height of Tower(BC) = 100 m
• Height of Flag ( AB) = 73 m

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