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From a window A, 10 m above the ground the angle of elevation of the top C of a tower is x°, where tan x° = 52\dfrac{5}{2} and the angle of depression of the foot D of the tower is y°, where tan y° = 14\dfrac{1}{4}. Calculate the height CD of the tower in metres.

From a window A, 10 m above the ground the angle of elevation of the top C of a tower is x°, where tan x° and the angle of depression of the foot D of the tower is y°, where tan y°. Calculate the height CD of the tower in metres. Heights and Distances, Concise Mathematics Solutions ICSE Class 10.

Heights & Distances

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Answer

From figure,

From a window A, 10 m above the ground the angle of elevation of the top C of a tower is x°, where tan x° and the angle of depression of the foot D of the tower is y°, where tan y°. Calculate the height CD of the tower in metres. Heights and Distances, Concise Mathematics Solutions ICSE Class 10.

⇒ AB = DE = 10 m.

In ∆AED

⇒tan y°=PerpendicularBase⇒tan y°=DEAE⇒14=DEAE⇒AE=4DE=4×10=40 m.\Rightarrow \text{tan y°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan y°} = \dfrac{DE}{AE} \\[1em] \Rightarrow \dfrac{1}{4} = \dfrac{DE}{AE} \\[1em] \Rightarrow AE = 4DE = 4 \times 10 = 40 \text{ m}.

In ∆AEC,

⇒tan x°=PerpendicularBase⇒tan x°=CEAE⇒52=CEAE⇒CE=AE×52=40×52=100 m.\Rightarrow \text{tan x°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan x°} = \dfrac{CE}{AE} \\[1em] \Rightarrow \dfrac{5}{2} = \dfrac{CE}{AE} \\[1em] \Rightarrow CE = AE \times \dfrac{5}{2} = 40 \times \dfrac{5}{2} = 100 \text{ m}.

From figure,

CD = DE + CE = 10 + 100 = 110 m.

Hence, height of tower (CD) = 110 m.

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