A pole 6m high costs a shadow 2-3m long on the ground then find the sun-s elevation-

R.E.F image #160;Given height of pole =6mlength of shadow = 2√(3)m Let ' θ ' #160;be elevation. of sum #160;then tanθ = AB/BC=6/2√(3) ...

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Standard X

Mathematics

Calculating Heights and Distances

A pole 6m h...

Question

A pole $6 m$ high costs a shadow $2 \sqrt{3} m$ long on the ground then find the sun's elevation..

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Solution

R.E.F image
Given height of pole =6m
length of shadow = $2 \sqrt{3} m$
Let ' $θ$ ' be elevation. of sum then
$t a n θ$ = $\frac{A B}{B C} = \frac{6}{2 \sqrt{3}}$
$t a n θ = \sqrt{3}$
$t a n θ = t a n \frac{π}{3}$
then $θ = \frac{π}{3}$
$∴$ Angle of elevation of sum is $60^{\circ} .$

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A pole 6m high costs a shadow 2√3m long on the ground then find the sun's elevation..

R.E.F image Given height of pole =6mlength of shadow =2√3mLet 'θ ' be elevation. of sum thentanθ =ABBC=62√3tanθ=√3tanθ=tanπ3 then θ=π3∴ Angle of elevation of sum is 60∘.

A pole $6 m$ high costs a shadow $2 \sqrt{3} m$ long on the ground then find the sun's elevation..

R.E.F image
Given height of pole =6m
length of shadow = $2 \sqrt{3} m$
Let ' $θ$ ' be elevation. of sum then
$t a n θ$ = $\frac{A B}{B C} = \frac{6}{2 \sqrt{3}}$
$t a n θ = \sqrt{3}$
$t a n θ = t a n \frac{π}{3}$
then $θ = \frac{π}{3}$
$∴$ Angle of elevation of sum is $60^{\circ} .$