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In Fig. 6.31, if PQ | ST, 2 POR - 110° and
ZRST = 130°, find ZQRS.
[Hint : Draw a line parallel to ST through
point R.]
R
Fig. 63
Answers
Step-by-step explanation:
Let us draw a parallel line XY to PQ || ST and passing through point R.
Sum of interior angle on the same side of the transversal is always = 180°
So that
∠ PQR + ∠ QRX = 180°
Given that ∠ PQR= 110°
110° + ∠QRX = 180°
∠QRX = 180° -110°
∠QRX = 70°
Sum of interior angle on the same side of the transversal is always = 180°
∠RST + ∠SRY = 180° (Co-interior angles on the same side of transversal SR)
Also
130° + ∠SRY = 180°
∠SRY = 50°
XY is a straight line. Use property of linear pair we get
∠QRX + ∠QRS + ∠SRY = 180°
70° + ∠QRS + 50° = 180°
∠QRS = 180° − 120°
= 60°
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Answer:
60 is the answer
Step-by-step explanation:
∠ PQR + ∠ QRX = 180°
Given that ∠ PQR= 110°
110° + ∠QRX = 180°
∠QRX = 180° -110°
∠QRX = 70°
Sum of interior angle on the same side of the transversal is always = 180°
∠RST + ∠SRY = 180° (Co-interior angles on the same side of transversal SR
130° + ∠SRY = 180°
∠SRY = 50°
XY is a straight line. Use property of linear pair we get
∠QRX + ∠QRS + ∠SRY = 180°
70° + ∠QRS + 50° = 180°
∠QRS = 180° − 120°