Question

Two buildings are facing each other on a road of width 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?
In the adioining figure twn cimles interont​

Answer 1

Answer:

Let AB be small building and CD be tall building

\therefore ED=BA=10\ m∴ED=BA=10 m

Let CE=xCE=x

In \triangle{BCE}△BCE

\tan60=\dfrac{CE}{BE}tan60=

BE

CE

\Rightarrow \sqrt3=\dfrac{CE}{12}⇒

3

=

12

CE

\Rightarrow CE=12\sqrt3⇒CE=12

3

\Rightarrow CE=20.78\ m⇒CE=20.78 m

Hence. CD=CE+EDCD=CE+ED

CD=20.78+10CD=20.78+10

CD=30.78\ mCD=30.78 m

Thus, height of the second building will be 30.78 m

Step-by-step explanation:

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