# 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 15 m tall is 205 m away from a tower 22 m high Determine the angle of elevation of the t...

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°

## Answer (Detailed Solution Below)

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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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