Question

in fig 13.6,POQ is a straight line. Find the value of x. also, find angle POR and angle QOS


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Answer 1

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∠POR=3x and ∠QOR=2x+10 [ Given ]

POQ is a line.

∴ ∠POR+∠QOR=180

o

[ Linear pair ]

⇒ 3x+2x+10

o

=180

o

⇒ 5x=170

o

⇒ x=

5

170

o

⇒ x=34

o

Answer 2

Given :

• ∠SOQ = x + 15
• ∠SOR = x
• ∠ROP = x + 6

What To Find :

• Value of x

Solution :

∠SOQ + ∠SOR + ∠ROP = 180°

(Linear Pair)

(x + 15)° + (x)° + (x + 6)° = 180°

x + 15° + x + x + 6° = 180°⠀⠀⠀

3x + 21 = 180°

3x = 180° - 21°

3x = 159°

x = 159 ÷ 3

x = 159/3

x = 53°

Hence, the value of x is 53°

• Measure of ∠POR = (x + 6)°
• 53° + 6° = 59°
• ∠POR = 59°

• Measure of angle ∠QOS = (x + 15)°
• 53° + 15 ° = 68°
• ∠QOS = 68°