Question

a man observes two vertical poles with are fixed opposite to each other either side of the road. If the width of the road is 90feet and heights of the 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 60° .find the heights of the poles​

Answer 1

Answer:

$\bigstar{\bold{Height\:of\:AB=30\sqrt{3}\:ft}}$

$\bigstar{\bold{Height\:of\:ED=60\sqrt{3}\:ft}}$

Step-by-step explanation:

$\Large{\underline{\underline{\sf{Given:}}}}$

• Width of road is 90 ft
• Heights of the pole are in the ratio 1 : 2
• Angle of elevation of poles = 60°

$\Large{\underline{\underline{\sf{To\:Find:}}}}$

• Heights of the poles

$\Large{\underline{\underline{\sf{Solution:}}}}$

→ Let the height of AB be x

→ Let the height of ED be 2x

→ Let CD = y, BC = 90 - y

→ Consider Δ ABC

  $tan\:60=\dfrac{AB}{BC}$

  $tan\:60=\dfrac{x}{90-y}$

  $\sqrt{3}=\dfrac{x}{90-y}----equation\:1$

  x = ( 90 - y) √3

→ Consider Δ EDC

   $tan\:60=\dfrac{ED}{CD}$

  $tan\:60=\dfrac{2x}{y}$

  $\sqrt{3} =\dfrac{2x}{y}-----equation\:2$

→ The LHS of equation 1 and 2 are equal. Therefore RHS must also be equal.

   $\dfrac{x}{90-y} =\dfrac{2x}{y}$

→ Cancelling x on both sides,

   $\dfrac{1}{90-y} =\dfrac{2}{y}$

  y = 180 - 2y

  3y = 180

    y = 60

→ Substitute the value of x in equation 2

  √3 = 2x/60

 60√3 = 2x

→ Hence height of ED = 2x = 60√3 ft

→ Height of AB = x = (60√3)/3 = 30√3 ft

$\boxed{\bold{Height\:of\:AB=30\sqrt{3}\:ft}}$

$\boxed{\bold{Height\:of\:ED=60\sqrt{3}\:ft}}$

$\Large{\underline{\underline{\sf{Notes:}}}}$

$sin\:A=\dfrac{opposite}{hypotenuse}$

$cos\:A=\dfrac{adjacent}{hypotenuse}$

$tan\:A=\dfrac{opposite}{adjacent}$

 

Answer 2

FINAL ANS - The height of the poles are 30√3 feet and 60√3 feet

★ TO FIND : HEIGHTS OF POLES

★ SOLUTION :

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 feet

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

h1 = BC√3 = 30√3 feet

∴ h2 = 2 * h1 = 2 * 30√3 = 60√3 feet

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