From the top of an 8 m high building, the angle of elevation of the top of a cable tower is $60°$ and the angle of depression of its foot is $45°$. Determine the height of the tower. (Take $\sqrt{3} =

Question

From the top of an 8 m high building, the angle of elevation of the top of a cable tower is $60°$ and the angle of depression of its foot is $45°$. Determine the height of the tower. (Take $\sqrt{3} = 1.732$).

Accepted Answer

Step 1: Let $AB$ = height of building $= 8$ m
Let $CD$ = height of tower $= h$ m
$\angle DBE = 60°$
$\angle ACB = \angle EBC = 45°$
$AC = BE = y$ (let)
Step 2: In right $	riangle ABC$,
$	an 45° = \frac{8}{AC}$
$1 = \frac{8}{AC}$
$\Rightarrow AC = 8$ m $\Rightarrow y = 8$ m
Step 3: In right $	riangle BDE$, $	an 60° = \frac{h - 8}{BE}$
$\sqrt{3} = \frac{h - 8}{y} \Rightarrow \sqrt{3}y = h - 8$
$\sqrt{3}(8) = h - 8$
Step 4: $h = 8\sqrt{3} + 8 = 8(\sqrt{3} + 1)$
$h = 8(1.732 + 1) = 8(2.732) = 21.856$ m
$	herefore$ Height of tower $= 21.856$ m

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