Question

the angle of elevation of the top of a building from the foot of the tower is 30 degree and the angle of elevation of the top of the tower from the foot of the building is 45 degree. If the tower is 30m high find the height of the tower​

Answer 1

Answer:

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• Angel of elevation from the top of a building from the foot of a tower is 30°
• Angle of elevation from the top of the tower to the foot of the building is 45°
• Height of tower = 30 m
• Height of the building = ?

$\displaystyle\underline{\bigstar\:\textsf{According to the given Question :}}$

• Assume the height of the tower to be x m

In ∆ABC

$\displaystyle\sf :\implies tan \ \theta = \dfrac{Opposite}{Adjacent}$

$\displaystyle\sf :\implies tan \ 45^{\circ} = \dfrac{AB}{BC}$

$\displaystyle\sf :\implies tan \ 45^{\circ} = \dfrac{30}{BC}$

$\displaystyle\sf :\implies 1 = \dfrac{30}{BC}$

$\displaystyle\sf :\implies BC = 30 \ m$

• We shall use the same in ∆DBC and find the value of DC and the length of DC will be equal to the height of the building

In ∆DBC

$\displaystyle\sf \dashrightarrow tan \ \theta = \dfrac{Opposite}{Adjacent}$

$\displaystyle\sf \dashrightarrow tan \ 30^{\circ} = \dfrac{DC}{BC}$

$\displaystyle\sf \dashrightarrow \dfrac{1}{\sqrt{3}} = \dfrac{DC}{30}$

$\displaystyle\sf \dashrightarrow 30\times 1 = DC \times \sqrt{3}$

$\displaystyle\sf \dashrightarrow \underline{\boxed{\sf DC = \dfrac{30}{\sqrt{3}} \ m}}$

$\displaystyle\therefore\:\underline{\textsf{ Height of the building is} \bf{\dfrac{30}{\sqrt{3}} \ m}}$