Question

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3. the ship is from the lighthouse. (V3 =1.73)
Two buildings are facing each other on a road of width 12 metre. From the
top of the first building, which is 10 metre high, the angle of elevation of the top
of the second is found to be 60°. What is the height of the second building?

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Answer with Solution!

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Answer 1

Answer:

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Let AB and CE are two buildings are facing each other on a road BD of width 12m as shown in figure. From the top of the first building , which is 10m high, the angle of elevation of the top of the second building CD is found to be 60°.

Let E is the point on CD in such a way that AB = ED = 10cm

Let CE = x m

Now, ∆CAE,

tan∠CAE = CE/AE [ see figure ]

tan60° = x/12 [ ∵AE = BD = 12m ]

√3 = x/12

x = 12√3 m

Hence, height of second building is (x + 10) = 12√3 + 10 = 12 × 1.7 + 10

= 20.4 + 10 = 30.4 m

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Answer 2

Answer:

□ 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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