Question

the angle of depression of the top and bottom of a tower as seen from the top of a
$60 \sqrt{3}$
m high cliff are 45° and 60° respectively . find the height of the tower

Answer 1

Let AE be the height of the tower

AE= AC+CE

BD=CE= 60√3 and BC=DE

From ΔABC
tan45= AC/BC
= 1=AC/BC
= AC/BC

Again, from ΔBCE
tan60= CE/BC
= √3= 60√3/BC
= √3×BC= 60√3
= BC= 60

Since, BC=AC
Then, AC=60

Now, AE= AC+CE
              = 60+ 60√3
              = 60(1+√3)
So, the height of the tower is 60(1+√3)m

Hope this helps you!!