Question

ab cd if apq=50° and prd = 130 find qpr​

Answer 1

Answer:

80°

Step-by-step explanation:

Given: ABIICD

APQ=50°

PRD=130°

In ABIICD,

APR=DRP=130° (Atlernate interior angles)

APR = APQ+QPR

130=50+QPR

130-50=QPR

80= QPR

Therefore QPR=80°

Answer 2

ANSWER✔

$\large\underline\bold{GIVEN,}$

$\sf\dashrightarrow \angle APQ=50\degree$

$\sf\dashrightarrow \angle PRD=130\degree$

$\large\underline\bold{TO\:FIND,}$

$\sf\dashrightarrow the\:value\:of\: \angle QPR$

$\large\underline\bold{SOLUTION,}$

$\red{\text{BY ANGLE SUM PROPERTY,}}$

$\sf\dashrightarrow \angle PRD + \angle PRQ= 180\degree$

$\sf\implies 130\degree + \angle PRQ = 180\degree$

$\sf\implies \angle PRQ=180-130$

$\sf\implies \angle PRQ= 50\degree$

$\bf{\boxed{\bf{ \star\:\: \angle PRQ=50\degree \:\: \star}}}$

NOW,

$\sf\therefore \angle APQ + \angle CQP=180\degree \:-----\boxed{by \:angle\:sum\:property.}$

$\sf\implies 50\degree+ \angle CQP=180\degree$

$\sf\implies \angle CQP=180-50$

$\sf\implies \angle CQP=130\degree$

$\large{\boxed{\bf{ \star\:\: \angle CQP=130\degree\:\: \star}}}$

NOW

$\sf\therefore in\: \triangle PQR,$

$\red{\text{BY ANGLE SUM PROPERTY OF A TRIANGLE,}}$

$\sf\dashrightarrow \angle PQR + \angle PRQ+ \angle QPR= 180\degree$

$\sf\implies 50\degree+50 \degree+\angle QPR=180\degree$

$\sf\implies 100\degree+\angle QPR = 180\degree$

$\sf\implies \angle QPR = 180-100$

$\sf\implies \angle QPR = 80\degree$

$\large{\boxed{\bf{ \star\:\:\angle QPR = 80\degree \:\: \star}}}$

$\large\underline\bold{THE\: VALUE\:OF\:\angle QPR \:\:IS\: 80\degree}$

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