Question

In the figure , AB and CD are parallel lines and transversal EF intersect them at P and Q . If < APR =250 , < RQC = 300 and < CQF = 650 , then find the measure of < RPQ and< PRQ.

Answer 1

Given :-

• AB and CD are parallel lines and transversal EF intersect them at P and Q .
• ∠APR =25°
• ∠RQC = 30°
• ∠CQF = 65°

To find :-

• ∠RPQ
• ∠PRQ

Construction :-

• Draw a line ST parallel to AB .

Solution :-

→ AB || ST

→ AB || CD

So,

→ CD || ST { Two lines parallel to the same line are parallel to each other. }

Now, AB || ST and EF is the transversal, So,

→ ∠APR = ∠PRT { Alternate interior angles .}

→ ∠PRT = 25° ----------- Eqn.(1)

Similarly, CD || ST and EF is the transversal,

→ ∠QRT = ∠RQC

→ ∠QRT = 30° ----------- Eqn.(2)

Adding Eqn.(1) and Eqn.(2), we get,

→ ∠PRT + ∠QRT = 25° + 30°

→ ∠PRQ = 55° (Ans.)

Now, FPE is a straight line, So,

→ ∠CQF + ∠CQR + ∠RQP = 180° .{ Angles forming a linear pair are supplementary.}

→ 65° + 30° + ∠RQP = 180°

→ ∠RQP = 180° - 95°

→ ∠RQP = 85°

Now, in ΔPRQ, we have :-

→ ∠RPQ + 55° + 85° = 180° {Angle sum Property.}

→ ∠RPQ + 140° = 180°

→ ∠RPQ = 180° - 140°

→ ∠RPQ = 40° (Ans.)

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