A pole 6 m high is fixed on the top of a
tower. The angle of elevation of the top
of the pole observed from a point P on
the ground is 60°. The angle of
depression of the point P from the top
of the tower is 45º. Find the height of
the tower.
[Use V3 = 1.732]
[CBSE 2012]
Answers
EXPLANATION.
The angle of elevation from top of the pole
observe from a point p on the ground = 60°.
The angle of depression of the point p from
top of the tower = 45°.
To find heights of the tower.
< APC = 60°. And <APB = 45°.
From right angled triangle <APB
Tan ø = p/b = perpendicular/base.
→ Tan 45° = AB/AP
→ 1 = H / AP
→ AP = H ........(1)
From right angled triangle <APC
Tan ø = p/b = Perpendicular/Base
→ Tan 60° = AC/AP
→ Tan 60° = AB + BC / AP
→ √3 = H + BC / AP
→ √3 = H + 6 / AP
→ AP = H + 6 / √3 .........(2)
From equation (1) and (2) we get,
→ H = H + 6 / √3
→ √3H = H + 6
→ √3H - H = 6
→ H ( √3 - 1 ) = 6
→ H = 6 / √3 - 1
→ H = 6 / √3 - 1 X √3 + 1 / √3 + 1
→ H = 6(√3 + 1) / 3 - 1
→ H = 2 ( √3 + 1 )
→ H = 2 X 2.732
→ H = 5.464
Heights of the tower = 5.464.
Answer:
EXPLANATION.
The angle of elevation from top of the pole
observe from a point p on the ground = 60°.
The angle of depression of the point p from
top of the tower = 45°.
To find heights of the tower.
< APC = 60°. And <APB = 45°.
From right angled triangle <APB
Tan ø = p/b = perpendicular/base.
→ Tan 45° = AB/AP
→ 1 = H / AP
→ AP = H ........(1)
From right angled triangle <APC
Tan ø = p/b = Perpendicular/Base
→ Tan 60° = AC/AP
→ Tan 60° = AB + BC / AP
→ √3 = H + BC / AP
→ √3 = H + 6 / AP
→ AP = H + 6 / √3 .........(2)
From equation (1) and (2) we get,
→ H = H + 6 / √3
→ √3H = H + 6
→ √3H - H = 6
→ H ( √3 - 1 ) = 6
→ H = 6 / √3 - 1
→ H = 6 / √3 - 1 X √3 + 1 / √3 + 1
→ H = 6(√3 + 1) / 3 - 1
→ H = 2 ( √3 + 1 )
→ H = 2 X 2.732
→ H = 5.464
Heights of the tower = 5.464.
