Question

In figure line PQ and RS intersect at O. If angle ROQ = 90 degree a:b=1:5 find the value of C​

Answer 1

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Required answer-:

∠POR=3∠ROQ (Given)

∠POR+∠ROQ=180° (Linear pair)

⇒3∠ROQ+∠ROQ=180°

⇒4∠ROQ=180°

⇒∠ROQ=$\frac{180°}{4}$ =45°

Now ∠SOQ=∠POR

=3∠ROQ (Ver. opp. ∠s)

=3×45°=135°

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Answer 2

Step-by-step explanation:

Angle QOS = Angle b ( vertically opposite angles)

let angle a be x then angle b = 5x

Now , a + b + 90 = 180 ( straight line angle=180)

x + 5x = 180 - 90

6x = 90 =》 x = 15 degree

Now, C + Angle QOS = 180 ( straight line)

C + b = 180 ( angle QOS=b)

C + 5×15 = 180 =》 C = 180 - 75 = 105 degree.