Question

a man observes to ventricle poles which are fixed opposite to each other on either side of the road is the width of the road is 90 feet and the height of the pole are in the ratio of 1 is to 2 alpha the angle of elevation of the top from the point between the line segment the foot of the poles on the road is 60 degree find the height of the poles​

Answer 1

Answer:

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 feet  ← height of the first pole

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