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In the adjoining figure, a man stands on the ground at a point A, which is on the same horizontal plane as B, the foot of the vertical pole BC. The height of the pole is 10 m. The man’s eye is 2 m above the ground. He observes the angle of elevation of C, the top of the pole, as x°, where tan x° = (25)\Big(\dfrac{2}{5}\Big). Calculate the distance AB in metres.

In the adjoining figure, a man stands on the ground at a point A, which is on the same horizontal plane as B, the foot of the vertical pole BC. The height of the pole is 10 m. Volume And Surface Area of solid RSA Mathematics Solutions ICSE Class 10.

Heights & Distances

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Answer

Let's take AD to be the height of the man, AD = 2 m.

From figure, BE = AD = 2 m.

Also,

CE = BC - BE = (10 - 2) = 8 m.

In ΔCED,

⇒tan⁡θ=perpendicularbase⇒tan⁡x∘=CEDE⇒25=8DE⇒DE=8×52⇒DE=402⇒DE=20 m.\Rightarrow \tan \theta = \dfrac{perpendicular}{\text{base}} \\[1em] \Rightarrow \tan x^{\circ} = \dfrac{CE}{DE} \\[1em] \Rightarrow \dfrac{2}{5} = \dfrac{8}{DE} \\[1em] \Rightarrow DE = \dfrac{8 \times 5}{2} \\[1em] \Rightarrow DE = \dfrac{40}{2} \\[1em] \Rightarrow DE = 20 \text{ m.}

From figure,

AB = DE = 20 m.

Hence, AB = 20 m.

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