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    an observer 1.5 metres tall is 20.5 metres away from a tower 22 metres high.determine the angle of elevation of the top of the tower from the eye of the observer.

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    Free solutions for Mathematics Exemplar Problems - class 10 Chapter 9 - Introduction to Trigonometry and Its Equations Introduction to Trigonometry and Its Equations - Exercise 8.3 question 10. These explanations are written by Lido teacher so that you easily understand even the most difficult concepts

    NCERT Exemplar Solutions Class 10 Mathematics Solutions for Introduction to Trigonometry and Its Equations - Exercise 8.3 in Chapter 8 - Introduction to Trigonometry and Its Equations

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    An observer 1.5 m tall is 20.5 m away from a tower 22 m high. Determine the angle of elevation of the top of the tower from the eye of the observer.

    Answer:

    Height of tower (TW) = 22 m [Given]

    Height of observer (AB) = 1.5 m [Given]

    Distance between foot of tower and observer (BW) = 20.5 m [Given]

    Let θ = ∠ of elevation of the observer at the top of the tower

    Now, TM = 22 m – 1.5 m = 20.5 m

    AM = 20.5 m

    ∴ tan θ = 20.5/20.5 = 1

    ⇒ tan θ = 1 ⇒ tan θ = tan 45° ⇒ θ = 45°

    Hence, the angle of elevation of the top of the tower from observer’s eye is 45°.

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    An observer 1.5 metres tall is 20.5 metres away from a tower 22 metres high. Determine the angle of elevation of the top of the tower from the eye of the observer.

    Click here👆to get an answer to your question ✍️ An observer 1.5 metres tall is 20.5 metres away from a tower 22 metres high. Determine the angle of elevation of the top of the tower from the eye of the observer.

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    An observer 1.5 metres tall is 20.5 metres away from a tower 22 metres high. Determine the angle of elevation of the top of the tower from the eye of the observer.

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    Updated on : 2022-09-05

    Solution Verified by Toppr

    Correct option is A)

    As the height of the Observer is 1.5

    AE=22−1.5=20.5 In △AED tanθ= ED AE ​ = 20.5 20.5 ​ =1⇒θ=45 o

    Therefore, answer is 45

    o

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    [Solved] An observer 1.5 m tall is 20.5 m away from a tower 22 m high

    Given: An observer length = 1.5 m A tower height = 22 m And the distance between an observer and a tower = 20.5 m Formula used: tanθ = perpendicular/

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    An observer 1.5 m tall is 20.5 m away from a tower 22 m high. The angle of elevation of the top of the tower from the eye of the observer is -

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    30° 45° 60° 75°

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    Detailed Solution

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    Given:

    An observer length = 1.5 m

    A tower height = 22 m

    And the distance between an observer and a tower = 20.5 m

    Formula used:

    tanθ = perpendicular/base

    Calculation:

    From the above figure, we can say the length of AB = the  length of DE = 20.5 m

    The angle of elevation of the top of the tower from the eye of the observer = ∠CDE = θ(let)

    And the length of CE = BC - EB

    ⇒ CE = 22 - 1.5 ⇒ CE = 20.5 m In ΔDEC: tanθ = CE/DE ⇒ tanθ = 20.5/20.5 ⇒ tanθ = 1 ⇒ tanθ = tan45° ⇒ θ = 45°

    ∴ The angle of elevation of the top of the tower from the eye of the observer is 45°. 

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