Question

PLEASE HELP ME TO SOLVE THIS QUESTION.......PLEASE.....

The angle of elevation of the top of a pillar at any point C in the ground is 15 degree.On walking 100 m towards the pillar the angle becomes 30 degree. Find the height of the pillar and it's distance from C.

Answer 1

i hope u understand
i think this is right answer

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Height\:of\:pillar=50\sqrt{3}\:m}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}}$

$\tt:\implies Angle \: of \: elevation( \theta) = 30 \degree$

$\tt:\implies Angle \: of \: elevation( \theta_{o} ) = 60 \degree$

$\tt: \implies Distance \: DC = 100 \: m$

$\red{\underline \bold{To \: Find:}}$

$\tt: \implies Height \: of \: pillar =?$

• Given Question

$\tt \circ \: Let \: height \: of \: pillar \:be \: x$

$\bold{In \: \triangle \: ABC : }$

$\tt: \implies tan \: \theta = \frac{p}{b}$

$\tt: \implies tan \:30 \degree = \frac{AB}{BC}$

$\tt: \implies \frac{1}{ \sqrt{3} } = \frac{x}{BD + DC}$

$\tt: \implies \frac{1}{ \sqrt{3} } = \frac{x}{BD+ 100}$

$\tt: \implies BD+ 100 = \sqrt{3} x$

$\tt: \implies BD = \sqrt{3} x - 100 - - - - - (1)$

$\bold{In \: \triangle \: ABD: }$

$\tt: \implies tan \: \theta_{o} = \frac{p}{b}$

$\tt: \implies tan \: 60 \degree = \frac{AB}{BD}$

$\tt: \implies \sqrt{3} = \frac{x}{ \sqrt{3} x - 100}$

$\tt: \implies 3x - 100 \sqrt{3} = x$

$\tt: \implies 3x - x = 100 \sqrt{3}$

$\tt: \implies 2x = 100 \sqrt{3}$

$\tt: \implies x = \frac{100 \sqrt{3} }{2}$

$\green{\tt: \implies x = 50 \sqrt{3} \: m}$

$\green{\tt \therefore Height \: of \: pillar \: is \: 50 \sqrt{3} \: m}$