Two buildings are in front of each other on a road of width 15 meters. From the top of the first building, having a height of 12 meter, the angle of elevation of the top of the second building is 30°.What is the height of the second building?
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□ Two buildings are in font of each other on a road of width ( base ) = 15 m.
□ Height of the first building ( perpendicular ) = 12 m
□ And the angle of elevation of the top of the second building from first building = 30°.
□ Height of the second building = ?
□ Let's solve this question ,
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□ Let's suppose the height of the another remaining building be " x "
□ Then height of second building = " x " + height of first building.
□ Height of second building = x + 12.
□ Base of the building = base at the top of building.
=> 15 = 15
According to the question ,
=> Tan30° = perpendicular / base
[ Value of Tan30° = 1 / √3 ]
=> 1 / √3 = x / 15
=> x√3 = 15
=> x = 15 / √3
□ Now , we have to multiply 15 /√3 by √3 / √3
=> x = ( 15 /√3 ) × (√3 / √3 )
=> x = ( 15√3 ) / 3
=> x = 5√3
Hence , height of the remaining building = 5√3
[ Value of √3 = 1.73 ]
Now ,
=> 5 × 1.73
=> 8.65
□ Now , total height of the building = x + height of the first building [ as both are parallel ].
=> 8.65 + 15 = 23.65
□ Hence , the total height of the another building is 23.65 m. or 5√3 + 12.
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Bhavanavindamuri:
gud answer^_^
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Answer:
Two buildings are in front of each other on a road of width 15 meters. From the top of the first building, having a height of 12 meter, the angle of elevation of the top of the second building is 30°.What is the height of the second building?
Answer:
Let AB and CD be two building, with
AB = 12 m
And angle of elevation from top of AB to top of CD = ∠CAP = 30°
Width of road = BD = 15 m
Clearly, ABDP is a rectangle
With
AB = PD = 12 m
BD = AP = 15 m
And APC is a right-angled triangle, In ∆APC
⇒ CP = 5√3 m
Also,
CD = CP + PD = (5√3 + 12) m
Hence, height of other building is (12 + 5√3 m).
Step-by-step explanation:
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