 Krishna
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Step 1:  Examine the given problem and draw a imaginary figure Height of the tower PQ = 30 m

The angle of elevation of the top of the building from the foot of the tower = 30\degree

The angle of elevation of the top of the tower from the foot of the building = 60 \degree

Height of the building AB = h say

Step 2:  Find the distance between the building and tower

NOTE: We know opposite side of the right angle triangle and we have to calculate the adjacent side.

So, use \tan \theta = \frac{opposite}{adjacent}

From the figure: Right triangle  PQB

\tan 60\degree = \frac{PQ}{BQ}

\tan 60\degree = \frac{30}{BQ}

\sqrt{3} = \frac{30}{BQ}                          \because \tan 60\degree = \sqrt{3}

BQ = \frac{30}{\sqrt{3}}......................(1)

Step 3: Find the height of the building

From the figure: right triangle ABQ

\tan 30\degree = \frac{AB}{BQ}

\frac{1}{\sqrt{3}} = \frac{h}{BQ}                         \because \tan 60\degree = \frac{1}{\sqrt{3}}

h \sqrt{3} = BQ

h \sqrt{3} = \frac{30}{\sqrt{3}}                      \because Equation (1)

h = \frac{30}{\sqrt{3} * \sqrt{3}}

h = \frac{30}{3}

h = 10 m

The height of the building = 10 m