 # 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° ### Course Features

• Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution