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Home » The angles of elevation of the top of a tower from two points on the ground at distances a and b meters from the base of the tower and in the same line are complementary. Prove that the height of the tower is √ab meter.
The angles of elevation of the top of a tower from two points on the ground at distances a and b meters from the base of the tower and in the same line are complementary. Prove that the height of the tower is √ab meter.
Class 10 | Exercise 22C | Heights and Distances | ICSE | Selina
The angles of elevation of the top of a tower from two points on the ground at distances a and b meters from the base of the tower and in the same line are complementary. Prove that the height of the tower is √ab meter.

Selina Solutions Concise Class 10 Maths Chapter 22 ex. 22(C) - 10

SOLUTION:

Let AB to be the tower of height h meters and let C and D be two points on the level ground such that BC = b meters, BD = a meters, ∠ACB = α, ∠ADB = β.

Given,

    \[\alpha \text{ }+\text{ }\beta \text{ }=\text{ }{{90}^{o}}\]

In ∆ABC,

    \[\begin{array}{*{35}{l}} AB/BC\text{ }=\text{ }tan\text{ }\alpha \\ h/b\text{ }=\text{ }tan\text{ }\alpha \text{ }\ldots ..\text{ }\left( i \right) \\ \end{array}\]

In ∆ABD,

    \[\begin{array}{*{35}{l}} AB/BD\text{ }=\text{ }tan\text{ }\beta \\ h/a\text{ }=\text{ }tan\text{ }\left( {{90}^{o}}~\text{ }-\alpha \right)\text{ }=\text{ }cot\text{ }\alpha \text{ }\ldots ..\text{ }\left( ii \right) \\ \end{array}\]

multiplying (i) by (ii), we get

    \[\begin{array}{*{35}{l}} \left( h/a \right)\text{ }x\text{ }\left( h/b \right)\text{ }=\text{ }1 \\ {{h}^{2}}~=\text{ }ab \\ =>\text{ }h\text{ }=\text{ }\surd ab\text{ }meter \\ \end{array}\]

Therefore, height of the tower is √ab meter.

i

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