Question

Question:-
Rohini saw an eagle on the top of a tree at an angle of elevation Of 61 ,while she was standing at the door of her house. She went on the terrace of the house so that she could see it clearly. The terrace was at a height of 4m. While observing the eagle from there the angle of elevation was 52. At what height from the ground was the eagle? Give Your answer up to nearest integer.
(tan 61 =1.80, Tan52 = 1.28 , tan29 = 0.55 , tan 38 = 0.78)

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Answer is 14cm
No Spam​

Answer 1

Trigonometry

The following are the tips and concept that can be use to find the solution:

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• Having a basic knowledge of Trigonometric ratios and Angles.
• Trigonometric ratios are sin, cos, tan, cot, sec, cosec.
• The standard angles of these trigonometric ratios are 0°, 30°, 45°, 60° and 90°.
• Relationship between sides and T ratios.

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Given : Rohini saw an eagle on the top of a tree at an angle of elevation Of 61 ,while she was standing at the door of her house. She went on the terrace of the house so that she could see it clearly. The terrace was at a height of 4m. While observing the eagle from there the angle of elevation was 52.

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To Determine : At what height from the ground was the eagle?

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Evaluation :

We've been given that, the height of the terrace is 4m. i.e. DE = 4cm.

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In right ∆ACD, we have;

→ tan(61°) = AC/CD

→ tan(61°) = AB + BC/CD [∵ AC = AB + BC]

→ tan(61°) = AB + 4/CD [∵ BC = DE = 4cm]

→ tan(61°) = AB + 4/CD ……(1)

Similarly, In right ∆ABE, we have;

→ tan(52°) = AB/BE

→ BE = AB/tan(52°)

→ CD = AB/tan(52°) [∵ BE = CD]

Now, substituting the value of CD in equation (1), we get:

→ tan(61°) = [(AB + 4)/(AB/tan(52°))]

→ tan(61°) = AB + 4 * tan(52°)/AB

→ tan(61°) * AB/tan(52°) = AB + 4

→ AB + 4 = tan(61°) * AB/tan(52°)

→ AB + 4 = 1.80 * AB/1.28 [Values of angles are given]

→ AB + 4 = 1.40 * AB [1.80/1.28 = 1.40]

→ AB + 4 = 7AB/5

→ 5(AB + 4) = 7AB

→ 5AB + 20 = 7AB

→ 5AB - 7AB + 20 = 0

→ 5AB - 7AB = -4

→ -2AB = -20

→ 2AB = 20

→ AB = 10

Therefore, the height of egal = AB + BC.

→ Height of egal = 10 + 4

→ Height of egal = 14m

Hence, the egal was at the height of 14m from the ground.

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Extra Information

1. Relationship between sides and T-Ratios.

• sin(θ) = Height/Hypotenuse
• cos(θ) = Base/Hypotenuse
• tan(θ) = Height/Base
• cot(θ) = Base/Height
• sec(θ) = Hypotenuse/Base
• cosec(θ) = Hypotenuse/Height

2. Square formulae.

• sin²(θ) + cos²(θ) = 1
• 1 + tan²(θ) = sec²(θ)
• 1 + cot²(θ) = cosec²(θ)

3. Reciprocal Relationship.

• sin(θ) = 1/cosec(θ)
• cos(θ) = 1/sec(θ)
• tan(θ) = 1/cot(θ)
• cot(θ) = 1/tan(θ)
• sin(θ)/cos(θ) = 1/cot(θ)
• cos(θ)/sin(θ) = 1/tan(θ)
• sin2(θ) = 2sin(θ)cos(θ)
• cos2(θ) = cos²(θ) - sin²(θ)

Answer 2

Answer:

hope it is help you please check the photo