In given fig. find x and y
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Akash98987:
x=77° y=50°
sorry x=50° u=77°
y=77°
Answers
Answered by
1
Angle y +50=180(linear pair)
Y=130
Angle Y+x=180(linear pair)
X=180-y
X=180-130 =50
Y=130
Angle Y+x=180(linear pair)
X=180-y
X=180-130 =50
y + 50 + < RPD = 180
its ok juz edit ur answer
ok
Answered by
0
given :
∠APQ = 50°
∠DRP = 127°
∠QPR = y
∠ PQR =x
⇒⇒ ∠QPR = ∠PRD
∠ y = ∠ 127° { OPP. INTERIOR ANGLES ARE EQUAL )
∴ y = 127°
⇒⇒ ∠APQ = ∠PQR
∠ x = ∠ 50 ° { OPP. INTERIOR ANGLES ARE EQUAL )
∴ x = 50°
⇒⇒ lets take < PRQ as " z "
127° + Z° = 180°
Z = 180° - 127° ( BY TRANSPOSING )
∴ Z = 53°
∠APQ = 50°
∠DRP = 127°
∠QPR = y
∠ PQR =x
⇒⇒ ∠QPR = ∠PRD
∠ y = ∠ 127° { OPP. INTERIOR ANGLES ARE EQUAL )
∴ y = 127°
⇒⇒ ∠APQ = ∠PQR
∠ x = ∠ 50 ° { OPP. INTERIOR ANGLES ARE EQUAL )
∴ x = 50°
⇒⇒ lets take < PRQ as " z "
127° + Z° = 180°
Z = 180° - 127° ( BY TRANSPOSING )
∴ Z = 53°
can u plz tell the value of angle PRQ
ya juz a sec
< PRQ = 127 + < PRQ
127 + < PRQ = 180
( ANGLE SUM PROPERTY )
< PRQ = 180 - 127
< PRQ = 180 - 127 = 53
and the sum of all three angle of triangle cannot be more than 180° and according to ur answer x=50° y=127° the sum of both makes 177° so is prq angle 3° or 53°
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