 Krishna
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Step 1:  According to the given measurements construct an imaginary figure. The two poles of equal heights are AB and PQ say

AB = PQ = H

A point between the poles  = D say

The distance between the two poles (AB & PQ) = BQ = 120 feet

If you say length of BD = h m

Then, the length of DQ = (120 - h) m

The angle of elevation

Step 2: Determine the height of the pole by using the trigonometric ratios.

FROM THE FIGURE:

Take right triangle ABD

\tan \theta = \frac{opposite}{adjacent} = \frac{AB}{BD}

\tan 60\degree = \frac{AB}{h}

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

H = h \sqrt{3}.....................(1)

Take right triangle  PQD

\tan 30\degree = \frac{PQ}{DQ}

\frac{1}{\sqrt{3}} = \frac{H}{120 - h}                \because \tan 30\degree = \frac{1}{\sqrt{3}}

H = \frac{120 - h}{\sqrt{3}} ....................(2)

Step 3:  Solve equation (1) and (2)

H = h \sqrt{3} .......................(1)

H = \frac{120 - h}{\sqrt{3}}.......................(2)

From equation (1)&(2) we can write

h\sqrt{3} = \frac{120 - h}{\sqrt{3}}

h\sqrt{3} * \sqrt{3} = 120 - h

3h + h = 120

h = \frac{120}{4}

h = 30

Step 4: Find the unknown lengths by substituting the 'h' value.

From equation (1)

H = h \sqrt{3}

H = 30\sqrt{3}

The length of the  DQ = 120 - h

DQ = 120 - 30

DQ = 90

Therefore, Height of the pole H = 30\sqrt{3}

The distances of the point from the poles are BD = 30 feet

DQ = 90 feet