Question

If the angle elevation of the tap of a summit is 45° and a person climbs at an inclination of 30° up to 1km. where the angle of elevation of top becomes 60° then the height of the summit is?

Answer 1

ANSWER.

The height of the summit is

$\sf \to \: h \: = \dfrac{ \sqrt{3} + 1 }{2} \: \: \: k.m$

$\sf \to \: \orange{{ \underline{step - \: by \: - step \: - explanation}}}$

$\sf \to \: angle \: of \: elevation \: of \: top \: of \: summit \: = 45 {}^{0} \\ \\ \sf \to \: person \: climb \: in \: the \: inclination \: of \: 30 {}^{0} upto \: 1 \: km \\ \\ \sf \to \: angle \: of \: elevation \: of \: top \: becomes \: = 60 {}^{0}$

$\sf \to \: in \: \triangle \: cdf \\ \\ \sf \to \: \sin(30 {}^{0} ) = \frac{z}{1} \\ \\ \sf \to \: \dfrac{1}{2} = \dfrac{z}{1} \\ \\ \sf \to \: z \: = \frac{1}{2} .......(1)$

$\sf \to \: \cos( 30 {}^{0} ) = \dfrac{y}{1} \\ \\ \sf \to \: \dfrac{ \sqrt{3} }{2} = \dfrac{y}{1} \\ \\ \sf \to \: y = \frac{ \sqrt{3} }{2}$

$\sf \to \: in \: \triangle \: abc \\ \\ \sf \to \: \tan(45 {}^{0} ) = \frac{h}{x + y} \\ \\ \sf \to \: x + y = h \\ \\ \sf \to \: x = h - y \\ \\ \sf \to \: x = h - \frac{ \sqrt{3} }{2} ......(2)$

$\sf \to \: in \: \triangle \: bde \\ \\ \sf \to \: \tan(60 {}^{0} ) = \frac{h - z}{x} \\ \\ \sf \to \: \sqrt{3} = \frac{h - z}{x} \\ \\ \sf \to \: \sqrt{3} x = h - z \\ \\ \sf \to \: \sqrt{3} (h - \frac{ \sqrt{3} }{2} ) = h - \frac{1}{2} \\ \\ \sf \to \: \sqrt{3} ( \frac{2h - \sqrt{3} }{ \cancel{2}} ) = \frac{2h - 1}{ \cancel{2}} \\ \\ \sf \to \: 2 \sqrt{3} - 3 = 2h - 1 \\ \\ \sf \to \: 2 \sqrt{3}h - 2h = 2$

$\sf \to \: \cancel{2}h( \sqrt{3} - 1) = \cancel{2} \\ \\ \sf \to \: h \: = \frac{1}{ \sqrt{3} - 1} \\ \\ \sf \to \: \frac{1}{ \sqrt{3} - 1} \times \frac{ \sqrt{3} + 1}{ \sqrt{3} + 1} \\ \\ \sf \to \: \frac{ \sqrt{3} + 1 }{3 - 1} = \frac{ \sqrt{3} + 1}{2} \\ \\ \sf \to \: \green{{ \underline{height \: of \: summit \: = \frac{ \sqrt{3} + 1 }{2} }}km}$