Question

. From the top of a 7m high building the angle of elevation of the top of a cable

tower is 600

and the angle of depression of its foot is 450

. Determine the height of

the tower.​

Answer 1

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:

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

Answer 2

$\huge\underline{\bf \orange{Aηsωeя :}}$

In ∆ADC,

tan(45° )= DC/DA

⟹ 1= DA7

⟹ DA=7m

In△ADE:

⟹tan(60 ∘ )= DE/DA

⟹tan(60 ∘ )= 7DE

⟹ 3 = 7DE

⟹ DE=7³

Height of tower= DC+ DE

Height of tower =7+7³

∴ Height of tower= 7(1+ 3)