Selina solutions

Selina solutions

Grade 6

Angles (With their Types) | Exercise 24(A)

Question 10

Q10)

In the given figure, AOB is a straight line. Find the value of x and also answer each of the following :

(i) ∠AOP = ……..

(ii) ∠BOP = ……..

(iii) which angle is obtuse ?

(iv) which angle is acute ?

Solution 10:

∠AOP = x + 30°

∠BOP = x- 30°

But, ∠AOP + ∠BOP = 180°

> x + 30° + x - 30° = 180°

> 2x = 180°

> x = 90°

(i) ∠AOP = x + 30° = 90° + 30° = 120°

(ii) ∠BOP = x - 30° = 90° - 30° = 60°

(iii) ∠AOP is an obtuse angle

(iv) ∠BOP is an acute angle

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subject-cta

Question 10

Q10)

In the given figure, AOB is a straight line. Find the value of x and also answer each of the following :

(i) ∠AOP = ……..

(ii) ∠BOP = ……..

(iii) which angle is obtuse ?

(iv) which angle is acute ?

Solution 10:

∠AOP = x + 30°

∠BOP = x- 30°

But, ∠AOP + ∠BOP = 180°

> x + 30° + x - 30° = 180°

> 2x = 180°

> x = 90°

(i) ∠AOP = x + 30° = 90° + 30° = 120°

(ii) ∠BOP = x - 30° = 90° - 30° = 60°

(iii) ∠AOP is an obtuse angle

(iv) ∠BOP is an acute angle

Still have questions? Our expert teachers can help you out

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subject-cta
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