The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Solution:
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Answers
Let AB and CD be the poles of equal height.
O is the point between them from where the height of elevation taken. BD is the distance between the poles.
As per above figure, AB = CD,
OB + OD = 80 m
Now,
In right ΔCDO,
tan 30° = CD/OD
1/√3 = CD/OD
CD = OD/√3 … (1)
Again,
In right ΔABO,
tan 60° = AB/OB
√3 = AB/(80-OD)
AB = √3(80-OD)
AB = CD (Given)
√3(80-OD) = OD/√3 (Using equation (1))
3(80-OD) = OD
240 – 3 OD = OD
4 OD = 240
OD = 60
Putting the value of OD in equation (1)
CD = OD/√3
CD = 60/√3
CD = 20√3 m
Also,
OB + OD = 80 m
⇒ OB = (80-60) m = 20 m
Thus, the height of the poles are 20√3 m and distance from the point of elevation are 20 m and 60 m respectively.
Answer:
Let AB and CD be the poles of equal height.
O is the point between them from where the height of elevation taken. BD is the distance between the poles.
As per above figure, AB = CD,
OB + OD = 80 m
Now,
In right ΔCDO,
tan 30° = CD/OD
1/√3 = CD/OD
CD = OD/√3 … (1)
Again,
In right ΔABO,
tan 60° = AB/OB
√3 = AB/(80-OD)
AB = √3(80-OD)
AB = CD (Given)
√3(80-OD) = OD/√3 (Using equation (1))
3(80-OD) = OD
240 – 3 OD = OD
4 OD = 240
OD = 60
Putting the value of OD in equation (1)
CD = OD/√3
CD = 60/√3
CD = 20√3 m
Also,
OB + OD = 80 m
⇒ OB = (80-60) m = 20 m
Thus, the height of the poles are 20√3 m and distance from the point of elevation are 20 m and 60 m respectively.