If the angle of elevation of the top of a hill from two points at distances a and b from the base and in the same straight line with it are complementary, then find the height of the hill.

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Solution

tan X = $\frac{h}{a} a n d t a n Y = \frac{h}{b}$
now tan (x + y) = $\frac{(t a n X + t a n Y)}{(1 - t a n X t a n Y)}$
Now since X + Y = 90 deg.
tan (X + Y ) = infinity
$\to$ Denominator = 0
$\to$ tan X tan Y = 1
$\to {\frac{h}{a}}^{*} \frac{h}{b} = 1$
$\to h^{2} = a b$
h = $\sqrt{a b}$

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If the angle of elevation of the top of a hill from two points at distances a and b from the base and in the same straight line with it are complementary, then find the height of the hill.

tan X = ha and tan Y=hb now tan (x + y) = (tan X+tan Y)(1−tan X tan Y) Now since X + Y = 90 deg. tan (X + Y ) = infinity → Denominator = 0 → tan X tan Y = 1 →ha∗hb=1 →h2=ab h = √ab

tan X = $\frac{h}{a} a n d t a n Y = \frac{h}{b}$
now tan (x + y) = $\frac{(t a n X + t a n Y)}{(1 - t a n X t a n Y)}$
Now since X + Y = 90 deg.
tan (X + Y ) = infinity
$\to$ Denominator = 0
$\to$ tan X tan Y = 1
$\to {\frac{h}{a}}^{*} \frac{h}{b} = 1$
$\to h^{2} = a b$
h = $\sqrt{a b}$