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```Chapter 4: Measurement of Angles – Exercise 4.1

Measurement of Angles – Exercise 4.1 – Q.1

(i)   9π/5

We have,

Now,

= 324°

(ii)   (-5π)/6

We have,

Now,

(iii)  (18π/5)c

We have

Now,

(iv) We have,

Now,

(v) We have,

Now,

= 630°

(vi) We have,

Now,

Measurement of Angles – Exercise 4.1 – Q.2

(i) 300°

We have, 180° = πc

Now,

(ii) 35°

We have,

180° = πc

Now,

(iii) – 56°

We have

180° = πc

Now,

(iv) 135°

We have,

180° = πc

Now,

(v)  – 300°

We have

180° = πc

Now,

(vi) 7°301

We have,

180° = πc

(vii) 125°301

We  have

180° = πc

(viii)  – 47°301

we have,

180° = πc

Measurement of Angles – Exercise 4.1 – Q.3

Let θ1 and θ2 be two acute angles of a right angled triangle.

∴ difference of acute angles

∵  in a right angled triangle,

On solving

From equation (ii)

So angles in degrees

Measurement of Angles – Exercise 4.1 – Q.4

Let θ1 and θ2 and θ3 be the angle or triangle.

Now,

We have to express all the angles in degrees

By angle slam property,

⟹ x = 40°

∴  θ1 = 24°, θ2 = 60°, θ3 = 96°
```

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