Question

solve questiones please​

Answer 1

Answer:

Hey Mate...!!!✌

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\bold{\Huge\red{\boxed{{{Answer}}}}}

Answer

➤ \: < /p > < p > < /p > < p > \angle{BPR}=70°➤</p><p></p><p>∠BPR=70°

➤ \: < /p > < p > < /p > < p > \angle{QPR}=80°➤</p><p></p><p>∠QPR=80°

\bold{\Huge\purple{\boxed{{{Given}}}}}

Given

➤ \: < /p > < p > < /p > < p > \angle{APQ}=50°➤</p><p></p><p>∠APQ=50°

➤ \: < /p > < p > < /p > < p > \angle{PRD}=130°➤</p><p></p><p>∠PRD=130°

➤ < /p > < p > < /p > < p > \: AB \: || \: CD➤</p><p></p><p>AB∣∣CD

\bold{\Huge\green{\boxed{{{To\: Find}}}}}

ToFind

➤ \: < /p > < p > < /p > < p > \angle{BPR}➤</p><p></p><p>∠BPR

➤ \: < /p > < p > < /p > < p > \angle{QPR}➤</p><p></p><p>∠QPR

\bold{\Huge\blue{\boxed{{{Explanation}}}}}

Explanation

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➠ As line AB || CD,

{\angle{APQ}=\angle{PQR}}\: {{ (Alternate Angles)}}∠APQ=∠PQR(AlternateAngles)

= > \angle{PQR} = 50°=>∠PQR=50°

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➠ On line QD,

= > \angle{PRD} + \angle{PRQ} = 180°=>∠PRD+∠PRQ=180°

= > 130° + \angle{PRQ} = 180°=>130°+∠PRQ=180°

= > \angle{PRQ} = 180° - 130°=>∠PRQ=180°−130°

= > \angle{PRQ} = 50°=>∠PRQ=50°

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➠ In ΔPQR,

\small{{= > \angle{PRQ} + \angle{PRD} + \angle{QPR} = 180° (Angle \: Sum \: Property \: of \: Δ)}}=>∠PRQ+∠PRD+∠QPR=180°(AngleSumPropertyofΔ)

= > 50° + 50° + \angle{QPR} = 180°=>50°+50°+∠QPR=180°

= > 100° + \angle{QPR} = 180°=>100°+∠QPR=180°

= > \angle{QPR} = 180° - 100°=>∠QPR=180°−100°

\boxed{\bold{= > \angle{QPR} = 80°}}

=>∠QPR=80°

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➠ On line AB,

\small{= > \angle{QPR} + \angle{APQ} + \angle{BPR} = 180° (Linear \: Pair)}=>∠QPR+∠APQ+∠BPR=180°(LinearPair)

= > 80° + 50° + \angle{BPR} = 180°=>80°+50°+∠BPR=180°

= > 130° + \angle{BPR} = 180°=>130°+∠BPR=180°

= > \angle{BPR} = 180° - 130°=>∠BPR=180°−130°

\boxed{\bold{= > \angle{BPR} = 50°}}

=>∠BPR=50°

Hope It Helps You....❤