Question

iv) 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 to
of the second is found to be 60°. What is the height of the second building?​

Answer 1

Answer:

let be present DC is secound building.

let present AB is first building.

and angle 60°

now,

tan 60°=CE/AE

=CE/12

$\sqrt{3}$

CE=12

$12 \sqrt{3}$

DC=DE+EC

$dc = 10 + 12 \sqrt{3}$

secound building height is

$10 + 12 \sqrt{3}$

Answer 2

Step-by-step explanation:

Answer in above attachment