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Two men standing on the same side of a tower in a straight line with it, measure the angles of elevation of the top of the tower as 25° and 50° respectively. If the height of the tower is 70 m, find the distance between the two men.

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Answer

Two men standing on the same side of a tower in a straight line with it, measure the angles of elevation of the top of the tower as 25° and 50° respectively. If the height of the tower is 70 m, find the distance between the two men. Volume And Surface Area of solid RSA Mathematics Solutions ICSE Class 10.

From figure,

CD is the distance between two persons.

In △ABC, we get

⇒tan⁡50∘=perpendicularbase=ABBC⇒tan⁡50∘=70x⇒x=70tan⁡50∘⇒x=701.1917⇒x=58.74 m.\Rightarrow \tan 50^{\circ} = \dfrac{\text{perpendicular}}{\text{base}} = \dfrac{AB}{BC} \\[1em] \Rightarrow \tan 50^{\circ} = \dfrac{70}{x} \\[1em] \Rightarrow x = \dfrac{70}{\tan 50^{\circ}} \\[1em] \Rightarrow x = \dfrac{70}{1.1917} \\[1em] \Rightarrow x = 58.74 \text{ m.}

In △ABD, we get

⇒tan⁡25∘=Perpendicularbase=ABBD⇒tan⁡25∘=70x+y⇒x+y=700.4663⇒x+y=150.12 m.\Rightarrow \tan 25^{\circ} = \dfrac{\text{Perpendicular}}{\text{base}} = \dfrac{AB}{BD} \\[1em] \Rightarrow \tan 25^{\circ} = \dfrac{70}{x + y} \\[1em] \Rightarrow x + y = \dfrac{70}{0.4663} \\[1em] \Rightarrow x + y = 150.12 \text{ m.}

CD = BD - BC = (x + y) - x = 150.12 - 58.74 = 91.38 m.

Hence, distance between the two men is 91.38 m.

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