Question

Two buildings are facing each other on a road of widht 12 metres. From the top of the first building which is 10 metres high, the angle of elevation of the top of the second is found to be 60°. What is the height of the second building?​

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

Answer 2

Step-by-step explanation:

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