Question

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 the top of the poles are 60° and 30° respectively. Find the height of the poles and the distance of the point from the poles.
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Class 10th ♥️
CBSE ♥️
Subject - Mathematics ♥️
Chapter 9 - Some Applications Of Trigonometry ♥️
Exercise 9.1 ♥️
Question number 10 ♥️

Answer 1

• The height (h) of the poles (AB and CD) = 20√3 m.
• The distance of point P from pole CD = CP = 20 m.
• The distance of point P from pole AB = BP = 60 m.

Given :

• Two poles of equal heights are standing opposite each other on either side of the road = BC = 80 m.
• From a point between them on the road, the angles of elevation of the top of the poles = 60° and 30°.

To Find :

• The height of the poles.

The distance of the point P from the poles :-

• The distance of point P from pole CD = CP.
• The distance of point P from pole AB = BP.

Solution :

Let,

AB and CD are poles of equal heights (h).

Point observer standing at Point P.

CP be x m.

BP be 80 - x m.

In ∆PCD,

We know that,

$\implies \bf tan \: \theta = \dfrac{P}{B}$

Where,

• P = Perpendicular = CD = h.
• B = Base = CP = x.

$\implies \bf tan \: {60}^{\circ} = \dfrac{CD}{PC}$

We know that,

• tan 60° = √3.

$\implies \bf \sqrt{3} = \dfrac{h}{x} \\ \\ \\ \implies \bf \sqrt{3} \: x = h \\ \\ \\ \: \: \therefore \: \: \: \bf h = \sqrt{3} \: x \: \: \: ..........(1)$

In ∆ABC,

$\implies \bf tan \: \theta = \dfrac{P}{B}$

Where,

• P = Perpendicular = AB = h.
• B = Base = BP = 80 - x.

$\implies \bf tan \: {30}^{\circ} = \dfrac{AB}{BP}$

We know that,

• tan 30° = 1 / √3.

$\implies \bf \dfrac{1}{ \sqrt{3} } = \dfrac{h}{80 - x} \\ \\ \\ \implies \bf \dfrac{80 - x}{ \sqrt{3} } = h \\ \\ \\ \: \: \therefore \: \: \bf h = \dfrac{80 - x}{ \sqrt{3} } \: \: \: ..........(2)$

Now, we need to find the value of x (CP).

From equation (1) and equation (2),

$\implies \bf \sqrt{3} \: x = \dfrac{80 - x}{ \sqrt{3} } \: m\\ \\ \\ \implies \bf \sqrt{3} \times \sqrt{3} \: x = 80 - x \: \: m\\ \\ \\ \implies \bf {(\sqrt{3})}^{2} \: x = 80 - x \: \: m\\ \\ \\ \implies \bf 3x = 80 - x \: \: m\\ \\ \\ \implies \bf 3x + x = 80 \: m\\ \\ \\ \implies \bf 4x = 80 \: m\\ \\ \\ \implies \bf x = \dfrac{ \not80}{ \not4} \: m\\ \\ \\ \implies \bf x = 20 \: m \\ \\ \\ \: \: \therefore \bf \: \: x = 20 \: m$

Now, we have to find the height of the poles = h = AB and CD.

From equation (1),

$\implies \bf h = \sqrt{3} \: x$

Substitute the value of x in the equation (1),

$\implies \bf h = \sqrt{3} \: (20 \: m) \\ \\ \\ \implies \bf h = \sqrt{3} \times 20 \: m \\ \\ \\ \implies \bf h = 20 \sqrt{3} \: m$

Hence, the height (h) of the poles (AB and CD) is 20√3 m.

Now, we have to find the distance of the point p from the poles (AB = BP and CD = CP).

Where,

$\implies \bf CP = x \\ \\ \\ \implies \bf CP = 20 \: m$

Therefore, CP is 20 m.

$\implies \bf BP = 80 - x \: m\\ \\ \\ \implies \bf BP = 80 - 20 \: m \\ \\ \\ \implies \bf BP = 60 \: m$

Therefore, BP is 60 m.

Hence,

The height (h) of the poles (AB and CD) = 20√3 m.

The distance of point P from pole CD = CP = 20 m.

The distance of point P from pole AB = BP = 60 m.

Answer 2

Answer:

Given :-

• 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 angle of elevation of the top of the poles are 60° and 30° respectively.

To Find :-

• What is the height of the poles and the distance of the poles.

Solution :-

• Let, AD and BC be the two poles of equal height (h) m.
• Let, P be the point on the road such that AP = x m and BP = (80 - x) m.
• ∠APD = 60° and ∠BPC = 30°.

➣ In ∆ APD, we have,

↦ tan 60° = $\dfrac{AD}{AP}$

↦ $\sqrt{3}$ = $\dfrac{h}{x}$

➟ x = $\dfrac{h}{\sqrt{3}}$ ...... equation no ❶

➣ Again, in ∆ BPC, we have,

↦ tan 30° = $\dfrac{BC}{BP}$

↦ $\dfrac{1}{\sqrt{3}}$ = $\dfrac{h}{80 - x}$

➾ By doing cross multiplication we get,

↦ 80 - x = $\sqrt{3}$ h

➟ x = 80 - $\sqrt{3}$ h ..... equation no ❷

➾ By comparing the equation no (1) and (2) we get,

↦ $\dfrac{h}{\sqrt{3}}$ = 80 - $\sqrt{3}$ h

↦ h = $\sqrt{3}$(80 - $\sqrt{3}$ h)

↦ h = 80$\sqrt{3}$ - 3 h

↦ h + 3 h = 80$\sqrt{3}$ - 3 h

↦ 4 h = 80$\sqrt{3}$

➟ h = 20$\sqrt{3}$

➾ By putting the value in the equation no (1), we get,

↦ x = $\dfrac{h}{\sqrt{3}}$

↦ x = $\sf\dfrac{20{\cancel{\sqrt{3}}}}{\cancel{\sqrt{3}}}$

➥ x = 20 m

» And, AP = x = 20 m

» Again, BP = 80 - x = 80 - 20 = 60 m

$\therefore$ The height of each pole is $\boxed{\bold{\small{20{\sqrt{3}}}}}$ and point P is at a distance of $\boxed{\bold{\small{20\: m}}}$ from left pole and from the right pole is $\boxed{\bold{\small{60\: m\: .}}}$