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. From the top of a 7m high building the angle of elevation of the top of a cabletower is 600and the angle of depression of its foot is 450. Determine the height of the tower.​

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254 cents Natali Aurelia
Answer:

Correct Question:

From the top of a 7m 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.

Given:

  • AB = BC = 7m
  • Angle CAD = 45°

♦To find:

  1. CE = ?

♠Method:

In ∆ADC,

\sf{\tan(45 {}^{ \circ} )=\dfrac{DC}{DA} } \\\\\\\implies \sf{1 =\frac{7}{DA} }\\\\\implies \sf{DA = 7m}

\sf{in\: \triangle \: ADE : } \\\\\implies \sf{ \tan(60 {}^{ \circ} )=\dfrac{DE}{DA} }\\\\\implies\sf{ \tan(60 {}^{ \circ} ) =\dfrac{DE}{7} } \\\\\implies \sf{ \sqrt{3}=\dfrac{DE}{7} } \\\\\implies \boxed{ \bf{DE= 7 \sqrt{3 } }}\\\\

 \sf{Height \: of \: tower= DC + DE} \\\\\implies \sf{Height \: of \: tower = 7 + 7 \sqrt{ 3} } \\\\\therefore \boxed{ \bf{Height \: of \: tower = 7(1 +\sqrt{3)} }}

Additional Information:

\bullet\:\sf Trigonometric\:Values :\\\\\boxed{\begin{tabular}{c|c|c|c|c|c}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & $\dfrac{1}{2} &$\dfrac{1}{\sqrt{2}} & $\dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & $\dfrac{\sqrt{3}}{2}&$\dfrac{1}{\sqrt{2}}&$\dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0&$\dfrac{1}{\sqrt{3}}&1&\sqrt{3}&Not D{e}fined\end{tabular}}

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Natali Aurelia
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