Question

a person standing on the bank of the river observes that the angle of elevation of the top of a tree on the opposite bank is 60°. when he retreats 20m from the bank,he observes the angle of elevation of the top of the same tree to be 30°. find the height of the tree and breadth of the river.​

Answer 1

Answer:

$\bigstar{\bold{Breadth\:of\:the\:river=10\:m}}$

$\bigstar{\bold{Height\:of\:the\:tree=10\sqrt{3}\:m}}$

Step-by-step explanation:

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

• The angle of elevation of the tree from the bank = 60°
• The angle of elevation 20 m away from the bank = 30°

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

• Height of the tree
• Breadth of the river

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

→ Let AB be the height of the tree

→ Let CB be the width of the river

→ Let DB be x

→ Hence CB = x - 20

→ Consider Δ ABD

  tan 30 = AB/DB

  tan 30 = AB/x

  1/√3 = AB/x

  x = AB√3 m

  AB = x√3/3 m-----(1)

→ Now consider Δ ABC

  tan 60 = AB/CB

  tan 60 = AB/x - 20

  √3 = AB/x-20

  AB = √3 (x - 20)------(2)

→ From equation 1 and 2, LHS are equal, hence RHS must be equal

  x√3/3 = √3 (x - 20)

→ Cancelling √3 on both sides,

  x/3 = x - 20

  x = 3x - 60

  x - 3x = -60

  -2x = -60

     x = 30

→ We know that width of the river = CB = x - 20 m

   CB = 30 - 20

   CB = 10 m

→ Hence breadth of the river is 10 m

$\boxed{\bold{Breadth\:of\:the\:river=10\:m}}$

→ Here AB = height of the tree

→ From equation 2,

  AB = √3 (30 - 20)

  AB = 10√3 m

→ Hence height of the tree is 10√3 m

$\boxed{\bold{Height\:of\:the\:tree=10\sqrt{3}\:m}}$

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

→ Sin A = opposite/hypotenuse

→ Cos A = adjacent/hypotenuse

→ Tan A = opposite/adjacent

 

 

Answer 2

Answer :-

✳ Height of the tree = 10√3 m

✳ Breadth of the river = 10 m

★Concept :-

• Here the concept used is of, Trigonometric ratios and heights and distances. In this case, what we do is, we have specific values for each angle in trigonometric ratios. We just apply those with the sides we received from practical application.

★ Solution :-

Given,

• Angle of elevation to the tree top from first position = <ACD = 60°

• Angle of elevation to the tree top after retreating 20 m = <ABD = 30°

• Distance between the final point of observation to the initial point = BC

= 20 m

▶Let the height of the tree be 'h' metres.

▶ Let the distance between the initial point of observation to the foot of tree be 'x' metres.

Then,

• AD = h metres

• CD = x metres

• Distance between the foot of thre tree and final point of observation = BD = (BC + CD)

✏ Now from right ∆ACD , we get,

•Tan < C = Perpendicular/Base = AD/CD

• Tan 60° = √3 (we know)

From this, we get,

✒ √3 = AD/CD

✒ AD = CD√3 .... (i)

✏ Now from right ∆ABD , we get,

• Tan < B = Perpendicular/Base = AD/BD

• Tan 30° = 1/√3 (we know)

From this we get,

✒ 1/√3 = AD / BD

✒ BD = AD√3 .... (ii)

From equation (i) and equation (ii), we get,

=> BD = AD√3

=> BD = (CD√3)√3

=> BD = 3(CD) .... (iii)

Then,

We know that,

BD = BC + CD

=> 3(CD) = 20 + x

And clearly, CD = x. So,

=> 3x = 20 + x

=> 2x = 20

=> x = 10 m

• Hence, Breadth of the river = x = 10 m

(since breadth of the river = distance between foot of the tree and initial point of observation)

Now,

=> AD = CD√3

=> AD = 10√3 m

• Hence, Height of the tree = h = 10√3 m

★ More to know :-

• Trigonometry - It is a branch of mathematics, which deals with calculations of sides and angles of triangle especially. It can be also used to point plots in graphs.

• Heights and Distances - These are the problems in which Trigonometry is used to find distance between two objects or their lengths.

• Ratio of Trigonometry involve sin, cosine, tangent, cotangent, cosec and sec.

=> sin ø = Perpendicular / Hypotenuse

=> cos ø = Base / Hypotenuse

=> tan ø = Perpendicular / Base

=> cosec ø = 1 / (sin ø)

=> sec ø = 1 / (cos ø)

=> cot ø = 1 / (tan ø)