∠BOC and ∠AOC are linear-pair-angles.

∠BOC + ∠AOC = 180

1) One of the angles forming a linear-pair is a right angle. What can you say about its other angle?

Let one of the angle forming a linear-pair be 'x' and other be y.

As ∠x = 90

We know that linear-pair-angles are supplementary.

∠x + ∠y = 180

90 + ∠y = 180

∠y = 180 - 90

∠y = 90

If one of the angles forming a linear pair is a right angle then other angle is also right angle.

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2) ∠PQR and ∠SQR are linear-pair-angles. If ∠PQR= 4x

and ∠SQR = 2x then find the value of x and measures of each angle.

As ∠PQR and ∠SQR form a linear-pair.

∴ ∠PQR + ∠SQR = 180

⇒ 4x + 2x = 180

⇒ 6x = 180

⇒ x = 30

m∠PQR = 4x = 4(30) = 120

m∠SQR = 2x =2(30) = 60

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3) ∠AOC = ∠COB, then show that ∠AOC = 90

Since ray OC stands on line AB.

∴∠AOC + ∠COB = 180 (Linear-Pair ) But ∠ AOC = ∠COB (given) ∴ ∠AOC + ∠AOC = 180 2∠AOC = 180 ⇒∠AOC = 90 ^{0} (Proved) |

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4) The two angles are in the ratio of 4:5. These two angles formed a linear-pair-angles. Find the measure of each.

Let the ratio be x.

So the two angles will be 4x and 5x.

We know that linear-pair-angles are supplementary.

4x + 5x = 180

9x = 180

x = 180 /9

x = 20

So, each angle will be ,4x = 4(20)= 80

Other angle = 5x = 5(20) = 100

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