 Krishna
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Step 1: According to the given data make a figure. NOTE: Height of the building AB = DE = h

The distance between the building and tower BD = AE = 7 m

The angle of elevation of the top of a cell tower = 60 \degree

The angle of depression to it's foot = 45 \degree

Step 2: Find the height of the cell tower by using the trigonometric ratios

Height of the cell tower DC = CE + ED

Determine the length of CE:

NOTE: From triangle ACE

\tan 60 \degree = \frac{opposite}{adjacent} = \frac{CE}{AE}

\tan 60 \degree = \frac{CE}{7}    (Since BD = AE)

\sqrt{3} = \frac{CE}{7} ( \because \tan 60 \degree = \sqrt{3} )

CE = x = 7 \sqrt{3} ..............(1)

Determine the length of ED:

From right triangle ABD

\tan 45\degree = \frac{opposite}{adjacent} = \frac{AB}{BD}

\tan 45\degree = \frac{AB}{7} (Since, equation (1))

1 = \frac{AB}{7} ( \because \tan 45\degree = 1 )

AB = 7

AB = DE = h = 7 m

Height of the  tower CD = CE + ED

CD = 7\sqrt{3} + 7

CD = 7(\sqrt{3} + 1)

CD = 7 (1.732 + 1)

CD = 7 (2.732)

CD = 19.124 m