Question

Two poles of equal heights are standing opposite to each other on either side of the
road, which is 100m wide. The angles of elevation of the top of the poles, from a point
between them on the road are 30 and 60°, respectively. Find the height of the poles
and the distances of the point from the poles. (See in figure)
​

Answer 1

Given:-

• The length of the road = 100 m
• The angle of elevation from a point on the ground to the top of both the poles = 30° and 60° respectively.

To Find:-

• The height of the poles [(AD) and (BC)]
• The distances of the point from the poles.

Assumption:-

• Let the point from where the angles of elevations to the top of the poles is raising be x

Note:-

• See the attachment for a better idea.

Solution:-

From the figure we have,

• ∠DXA = 30°
• ∠CXB = 60°
• AB = 100m

Now,

Since we have AB = 100 m

Hence,

AX = AB - BX

=> AX = 100 - BC

Again,

BX = AB - AX

=> BX = 100 - AX .............. (a)

Now,

In ∆ABX

∠AXB = 30°

AD = h

AX = 100 - BX ................ (b)

We know,

$\sf{tan30^\circ = \dfrac{1}{\sqrt{3}}}$

Hence,

$\sf{Tan30^\circ = \dfrac{AD}{AX}}$

$\sf{\implies \dfrac{1}{\sqrt{3}} = \dfrac{h}{100 - BX}}$

$\sf{\implies \dfrac{100-BX}{\sqrt{3}} = h}$

$\sf{\implies h = \dfrac{100-BX}{\sqrt{3}} \longrightarrow[i]}$

Now,

In ∆CBX,

CXB = 60°

BX = 100 - AX

BC = h

We know,

$\sf{tan60^\circ = \sqrt{3}}$

Hence,

$\sf{tan60^\circ = \dfrac{BC}{100-AX}}$

$\sf{\implies \sqrt{3} = \dfrac{h}{100-AX}}$

$\sf{\implies \sqrt{3}(100-AX) = h}$

$\sf{\implies h = \sqrt{3}(100-AX)\longrightarrow[ii]}$

From Equation [i] and Equation [ii]

$\sf{h = h}$

$\sf{\implies \dfrac{100 - BX}{\sqrt{3}} = \sqrt{3}(100 - AX)}$

$\sf{\implies 100 - BX = \sqrt{3}\times \sqrt{3}(100 - AX)}$

$\sf{\implies 100 - BX = 3(100 - AX)}$

From equation (b) we can write AX = 100 - BX

Hence,

$\sf{\implies 100 - BX = 3[100 - (100 - BX)]}$

$\sf{\implies 100 - BX = 3[100 - 100 + BX]}$

$\sf{\implies 100 - BX = 3\times BX}$

$\sf{\implies 100 - BX = 3BX}$

$\sf{\implies 100 = 3BX + BX}$

$\sf{\implies 100 = 4BX}$

$\sf{\implies BX = \dfrac{100}{4}}$

$\sf{\implies BX = 25\:m}$

Now,

Putting the value of BX in equation [i]

$\sf{h = \dfrac{100 - BX}{\sqrt{3}}}$

$\sf{h = \dfrac{100 - 25}{\sqrt{3}}}$

$\sf{h = \dfrac{75}{\sqrt{3}}}$

Rationalizing the denominator,

$\sf{h = \dfrac{75}{\sqrt{3}}\times \dfrac{\sqrt{3}}{\sqrt{3}}}$

$\sf{h = \dfrac{75\sqrt{3}}{3}}$

$\sf{\implies h = 25\sqrt{3}\:m}$

Therefore the height of the poles is 25√3 m.

Now,

We need to find the distances of the point from the poles.

We have,

AX = 100 - BX

Putting the value of BX,

AX = 100 - 25

=> AX = 75 m

Also we have,

BX = 100 - AX

Putting the value of AX

=> BX = 100 - 75

=> 25 m

Hence the distances of the point from the poles is 75m and 25 m respectively.

______________________________________

Answer 2

$\\ \blue{\underline{ \large \rm\color{blue} \: Solution :- }}$

Let two poles AB & CD ,

So ,length of pole = AB = CD

Also, length of the road = 100 cm

So, BC = 100 cm ,

Let's point P be a point on the road between the poles ,

We need to find the height of the poles i.e. AB & CD ,

And the distance of the point from the pole i.e BP & CP ,

Since , poles are perpendicular to the ground,

$\\ \implies \: \: \: \: \: \: \: \: \sf \: \angle \: \: ABP \: = 90 \: \degree \: \: and \: \: \angle \: DCP \: = \: 90 \degree$

$\\ \: \: \: \: \sf in \: right \: angled \: triangle \: \: DPC \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: tan \: p \: = \frac{side \: \: opposite \: to \: p}{side \: \: adjacent \: \: to \: \: p}$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \: \: \: \: \: \: \: \: \sf \: tan \: 30 \degree \: = \frac{ DC}{CP }$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \: \: \: \: \: \: \: \: \sf \: \frac{1}{ \sqrt{3} } \: = \frac{ DC}{CP }$

$\\ \: \: \: \: \: \: \: \: \: \: \: : \implies \: \: \: \: \: \: \: \: \sf \: \frac{CP}{ \sqrt{3} } \: = { DC}$

$\\ \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: : \implies \: \: \: \: \: \: \: \: \sf \: DC = \frac{CP}{ \sqrt{3} } \: .........................(1)$

________________________________

Now,

$\\ \: \: \: \: \sf in \: right \: angled \: triangle \: \: APB \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: tan \: p \: = \frac{side \: \: opposite \: to \: p}{side \: \: adjacent \: \: to \: \: p}$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \: \: \: \: \: \: \: \: \sf \: tan \: p \degree \: = \frac{ AB }{PB }$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \: \: \: \: \: \: \: \: \sf \: tan \: 60 \degree \: = \frac{ AB }{PB }$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \: \: \: \: \: \: \: \: \sf \: { \sqrt{3} } \: BP = \: = AB$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \therefore \: \: \: \: \: \: \: \: \: \sf \: CD = { \sqrt{3} } \: BP \: \: ...........(as \: ab \: = CD).....(2)$

From 1 & 2 ,

$\\ \: \: \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{CP }{ \sqrt{3} } = \sqrt{3} \: BP$

$\\ \: \: \sf \: \: \: \: \: \: : \implies \: \: \: \: \: \: \: \: \: {CP } = { \sqrt{3} } \times \sqrt{3} \: BP$

$\\ \: \: \sf \: \: \: \: \: \: : \implies \: \: \: \: \: \: \: \: \: {CP } = { {3} } \: BP$

Now,

BC = BP CP

⟶ 100 = BP + CP

⟶ 100 = BP + 3BP

⟶ 100 = 4BP

⟶ 4BP = 100

⟶ BP = 100/4

⟶ BP = 25cm

Now,

CP = BC - BP

⟶ CP = 100 - 25

⟶ CP = 75 m

From 2 ,

CD = √3BP

⟶ CD = √3 × 25

⟶ CD = 25√3

Hence , the length the length of the pole = CD = 25√3 ,

Hence the distances of the point from the poles is 75m and 25m respectively.