Math, asked by aseef, 10 months ago

A man observes two verticle poles which are fixed to each other on either side of the road.if width of road is 90feet and heights of pole are in the ratio 1:2,also the angle of elevation of their tops from a point between the line joining the foot of the poles on the road is 60degree.find height of poles.

Answers

Answered by bhagyashreechowdhury
0

The height of the poles are 30√3 feet and 60√3 feet.

Step-by-step explanation:

Referring to the figure attached below, let's make some assumptions

AB = h₁ = height of the first pole

ED = h₂ = height of the second pole

BD = distance between the two poles = 90 feet

Angle of elevation from point C to the top of AB = θ₁ = 60°

Angle of elevation from point C to the top of ED = θ₂ = 60°

Let the distance of point C from the foot of AB be “BC”, then the distance of point C from the foot of ED will be “CD = (90 - BC)”.

Since it is given that the ratio of the heights of the pole are 1:2

So, if the height of the first pole AB is “h1” then the height of the second pole ED will be "h2 = 2h1”.

Now,  

Consider ΔABC, applying the trigonometric ratios of a triangle, we get

tan θ₁ = perpendicular/base

⇒ tan 60° = AB/BC

⇒ √3 = h₁/BC

h₁ = BC√3 ...... (i)

and,

Consider ΔEDC, applying the trigonometric ratios of a triangle, we get

tan θ₂ = perpendicular/base

⇒ tan 60° = ED/CD

⇒ √3 = h₂/(90 - BC)

⇒ 2h₁ = √3 [90 - BC]

h₁= (√3/2) [90 - BC] ...... (ii)

From (i) & (ii), we get

BC√3 = (√3/2) [90 - BC]

⇒ 2BC = 90 – BC

⇒ 2BC + BC = 90

⇒ 3BC = 90

⇒ BC = 90/3

BC = 30

Substituting the value of BC in (i), we get

h₁ = BC√3 = 30√3  feetheight of first pole

h₂ = 2 * h₁ = 2 * 30√3 = 60√3  feet ← height of the second pole

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