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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, π radians = 180° Now, = 324° (ii) (-5π)/6 We have, π radians = 180° Now, (iii) (18π/5)c We have π radians = 180° Now, (iv) We have, Now, (v) We have, π radians = 180° Now, = 630° (vi) We have, π radians = 180° 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°
(i) 9π/5
We have,
π radians = 180°
Now,
= 324°
(ii) (-5π)/6
(iii) (18π/5)c
We have
(iv) We have,
(v) We have,
= 630°
(vi) We have,
(i) 300°
We have, 180° = πc
(ii) 35°
180° = πc
(iii) – 56°
(iv) 135°
(v) – 300°
(vi) 7°301
(vii) 125°301
(viii) – 47°301
we have,
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
Let θ1 and θ2 and θ3 be the angle or triangle.
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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