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
