A man observes two vertical poles which are fixed opposite to each other on either side of the road if the width of a road is 90 feet and heights of pole are in ratio 1:2 ,also the angle of elevation of their tops from points between the line joining foot of the poles on road is 60°. Find the height of poles
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Answer:
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Step-by-step explanation:
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Answer:
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Step-by-step explanation:
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
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