Question

10. Two poles of equal heights are standing opposite each other on either side of the road.
which is 80 m wide. From a point between them on the road, the angles of elevation of
of the poles are 60° and 30°, respectively. Find the height of the poles and the
distances of the point from the poles.
the top​

Answer 1

Hope this helps you

If it helps you pls mark as brainliest

Answer 2

${\bold{\pink{\underline{\green{G}\purple{iv}\orange{en}\red{:-}}}}}$

let AB and CD be the two poles of equal height.

• BD = 80 m

• ∠ AOB = 60°

• ∠ COD = 30°

${\bold{\pink{\underline{\pink{To}\:\purple{Fin}\blue{d}\red{:-}}}}}$

• AB and CD =?

• BO and OD = ?

${\bold{\pink{\underline{\red{So}\purple{lut}\green{ion}\orange{:-}}}}}$

AB = CD (given)

In ∆ ABO

$\bf\implies\:\dfrac{AB}{BO}=\dfrac{P}{B}=tan60\degree$

$\bf\implies\:\dfrac{AB}{BO}=\sqrt{3}$

$\bf\implies\:BO=\dfrac{AB}{\sqrt{3}}$

In ∆ CDO

$\bf\implies\:\dfrac{CD}{DO}=\dfrac{P}{B}=tan30\degree$

$\bf\implies\:\dfrac{CD}{DO}=\dfrac{1}{\sqrt{3}}$

$\bf\implies\:\dfrac{CD}{80-BO}=\dfrac{1}{\sqrt{3}}$

$\bf\implies\:CD\:\sqrt{3}=80-BO$

$\bf\implies\:AB\:\sqrt{3}=80-\dfrac{AB}{\sqrt{3}}$

$\bf\implies\:AB\:\sqrt{3}+\dfrac{AB}{\sqrt{3}}=80$

$\bf\implies\:AB\:\dfrac{\sqrt{3}}{1}+\dfrac{1}{\sqrt{3}}=80$

$\bf\implies\:AB\:\dfrac{3+1}{\sqrt{3}}=80$

$\bf\implies\:AB=\dfrac{4}{\sqrt{3}}=80$

$\bf\implies\:AB=\dfrac{80}{4}\times\:\sqrt{3}$

$\bf\implies\pink{{AB=20\sqrt{3}}}$

$\bf\blue{{AB=CD=20\sqrt{3}}}$

Now,

$\bf\:BO=\dfrac{AB}{\sqrt{3}}$

$\bf\:BO=\dfrac{20\sqrt{3}}{\sqrt{3}}$

$\bf\green{{BO=20\:m}}$

then,

$\bf\:OD=80-BO$

$\bf\:OD=80-20$

$\bf\red{{OD=60\:m}}$

Hence, Height of poles $\bf\orange{{= 20\sqrt{3}}}$

Distance of point 20 m and 60 m