Question

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Answer 1

$\huge\bf\star\fbox\red{Question:-}$$</p><p>\huge\bold\pink\bigstar{QUESTION }$

GIVEN IN QUESTION =PQ parallel to ST

<PQR=110°

and <RST=130°

we have to find<QRS

${\huge \fbox \pink {A}\fbox \blue {n} \fbox \purple {s} \fbox \green{w} \fbox \red {e} \fbox \orange {r}}$

$\huge\bold\pink\bigstar{Answer}$

AS WE COME TO KNOW

THAT < PQ PARALLEL TO <ST

AS COMPLARE ANGLE AS <1 ;<2;<3

THEN GIVEN <1=<2

<2=130° (IN QUESTION IT IS GIVEN )

AND

<2+<3=180°

because of linear pair equations

=130°+<3=180° (linear pair)

=<3=180°-130°=50°

then

<5+<4= (as we know 5 +4 =9)

then <9 =180°

110+<4=180°

<4=180°-110=70°

in TRIANGLE QMR

<3+<4+<R=180°

120°+<R=180°

<R=180°-120°

<R=60°

$\huge\bf\star\fbox\red{ANSWER:-}$[tex]

NOW WE COME TO KNOW THAT

<QRS

60 DEGREE

HERE IS UR ANS MATE

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Answer 2

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