Question

The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 60°, and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the Distance between the two towers and also the height of the other tower.​

Answer 1

Answer:

Hope this is OK for you completely

Answer 2

Answer:

let first tower be T1 and second tower be T2

so, height of T1 is 30m

Given:

angle of elevation of top of T1 from foot of T2=60°

angel of elevation of top of T2 from foot of T1=30°

To find out:

distance between T1 and T2

height of T2

$abc \: is \: a \: right \: angle \: triangl \\ ab = perpendicular = 30m \\ angle \: c = 60 \: degree \\ \tan(60) = \frac{perpendicular}{base} \\ \sqrt{3} = \frac{30}{base} \\ base = \frac{30}{ \sqrt{3} } \\ \\ triangle \: dbc \: is \: a \: right \: angle \: triangle \: so \\ \tan(30) = \frac{perpendicular}{base} \\ \frac{1}{ \sqrt{3} } = \frac{p}{ \frac{30}{ \sqrt{3} } } \\ \frac{1}{ \sqrt{3} } = \frac{p \sqrt{3} }{30} \\ 30 = 3p \\ p = 10$

So, the distance between T1 and T2 is 30\√3m

and height of T2 is 10m