Question

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?
[3 Marks]​

Answer 1

30.7 m

Step-by-step explanation:

height of first building (smaller)

= 10 m(given)

height of 2nd building (longer)

=( 10 + x )

width (horizontal distance between 2 buildings)

= 12m (given )

tan 60° = p/b = x / 12

√3= x /12 => x = 12√3 m

height of 2nd building= 10+12√3

= 10+ 12× 1.732 = 30.7 m

Answer 2

Hey mate,

30 . 78 metre

Let AB be small building and CD be small building .

.•. ED = BA = 10 m

Let CE = x

In △BCE

CE

Tan 60 = ––

BE

CE

----

===> √3 = 12

===> CE = 12 √3

Hence, CD = CE + ED

CD = 20 . 78 + 10

CD = 30 . 78

Thus, height of the second building will be

30.78 m

Hope it helps..

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