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 distances of the point from the poles.​

Answer 1

Answer:

The distance from the point of elevation are 20 m and 60 m respectively, the height of the poles is $\sf{20 \sqrt{3}}$

Step-by-step explanation:

Given :

Width of the road = 80 m

Angles of elevation = 60° and 30°

To find :

The height of the poles and the distance of the point from poles

Solution :

Consider -

• AB and CD = poles of equal height
• O point between them = point from where height of elevation taken
• BD = distance between the poles

$\rule{300}{1.5}$

Refer the given figure, as in the figure -

• AB = CD
• OB + OD = 80 m

In $\triangle$ COD (Right triangle) :

$\sf{\longrightarrow} \: \tan(30^{\circ}) = \dfrac{CD}{OD} \\ \\ \sf{\longrightarrow} \: \dfrac{1}{ \sqrt{3}} = \frac{CD}{OD} \\ \\ \sf{\longrightarrow} \: cd = \dfrac{OD}{ \sqrt{3}} \: \: ...... \: (1)$

$\rule{300}{1.5}$

In Right $\triangle$ ABO,

$\sf{\longrightarrow} \: \tan(60^{\circ} ) = \dfrac{AB}{OB} \\ \\ \sf{\longrightarrow} \: \sqrt{3} = \dfrac{AB}{(80 - OD)} \\ \\ \sf{\longrightarrow} \: AB = \sqrt{3}(80 - OD)$

$\rule{300}{1.5}$

It is given that AB = CD,

$\sf{\longrightarrow} \: \sqrt{3} (80 - OD) = \dfrac{OD}{\sqrt{3}} \\ \\ \sf{\longrightarrow} \: 3(80 - OD) = OD \\ \\ \sf{\longrightarrow} \: 240 - 3OD = OD \\ \\ \sf{\longrightarrow} \: 3OD + OD = 240 \\ \\ \sf{\longrightarrow} \: 4OD = 240 \\ \\ \sf{\longrightarrow} \: OD = \dfrac{240}{4} \\ \\ \sf{\longrightarrow} \: OD = 60$

$\rule{300}{1.5}$

Substitute the value of OD in equation 1,

$\sf{\longrightarrow} \: CD = \dfrac{OD}{ \sqrt{3}} \\ \\ \sf{\longrightarrow} \: CD = \dfrac{60}{ \sqrt{3} } \\ \\ \sf{\longrightarrow} \: CD = 20 \sqrt{3} \: \: m$

$\rule{300}{1.5}$

It is also given that OB + OD = 80m,

$\sf{\longrightarrow} \: OB + 60= 80 \\ \\ \sf{\longrightarrow} \: OB = 80 - 60 \\ \\ \sf{\longrightarrow} \: OB = 20$

OB = 20 m

$\therefore$ The distance from the point of elevation are 20 m and 60 m respectively, the height of the poles is $\sf{20 \sqrt{3}}$

Answer 2

Step-by-step explanation:

Given:-

AB and CD be the two poles of equal height.

Their heights be H m.

BC be the 80 m wide road.

P be any point on the road.

Let ,

CP be x m,

BP = (80 – x) . 

Also, ∠APB = 60° and ∠DPC = 30°

In right angled triangle DCP, 

Tan 30° = CD/CP 

⇒ h/x = 1/√3 

⇒ h = x/√3 ---------- (1) 

In right angled triangle ABP

Tan 60° = AB/AP 

⇒ h/(80 – x) = √3

⇒ h = √3(80 – x) 

⇒ x/√3 = √3(80 – x) 

⇒ x = 3(80 – x) 

⇒ x = 240 – 3x

⇒ x + 3x = 240

⇒ 4x = 240

⇒ x = 60 

Height of the pole, h = x/√3 = 60/√3 = 20√3. 

Thus, position of the point P is 60 m from C and height of each pole is 20√3 m.