Selina solutions

Selina solutions

Grade 6

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

Question 11

Q11)

In the given figure, PQR is a straight line. Find x. Then complete the following:

(i) ∠AQB = ……..

(ii) ∠BQP = ……..

(iii) ∠AQR = …….

Solution 11:

PQR is a straight line

∠AQP = x + 20°

∠AQB = 2x + 10°

∠BQR = x - 10°

But ∠AQP + ∠AQB + ∠BQR = 180°

> x + 20° + 2x + 10° + x - 10° = 180°

> 4x + 20° = 180°

> 4x = 180° - 20° = 160°

> x =

\frac{160}{4}=40^0

(i) ∠AQB = 2x + 10° = 2 x 40° + 10° = 80° + 10° = 90°

∠AQP = x + 2(T = 40° + 20° = 60°

∠BQR = x - 10° = 40° - 10° = 30°

(ii) ∠BQP = ∠AQP + ∠AQB = 60° + 90° = 150°

(iii) ∠AQR = ∠AQB + ∠BQR = 90° + 30° = 120°

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subject-cta

Question 11

Q11)

In the given figure, PQR is a straight line. Find x. Then complete the following:

(i) ∠AQB = ……..

(ii) ∠BQP = ……..

(iii) ∠AQR = …….

Solution 11:

PQR is a straight line

∠AQP = x + 20°

∠AQB = 2x + 10°

∠BQR = x - 10°

But ∠AQP + ∠AQB + ∠BQR = 180°

> x + 20° + 2x + 10° + x - 10° = 180°

> 4x + 20° = 180°

> 4x = 180° - 20° = 160°

> x =

\frac{160}{4}=40^0

(i) ∠AQB = 2x + 10° = 2 x 40° + 10° = 80° + 10° = 90°

∠AQP = x + 2(T = 40° + 20° = 60°

∠BQR = x - 10° = 40° - 10° = 30°

(ii) ∠BQP = ∠AQP + ∠AQB = 60° + 90° = 150°

(iii) ∠AQR = ∠AQB + ∠BQR = 90° + 30° = 120°

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subject-cta
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