In Fig. 6.31, if PQ ST, PQR = 110° and RST = 130°, find QRS.
[Hint : Draw a line parallel to ST through point R.]
Answers
Answer:
Given: PQ || ST, ∠PQR = 110° and ∠RST = 130°
To Find: ∠QRS
Lines which are parallel to the same line are parallel to each other.
When two parallel lines are cut by a transversal, co-interior angles formed are supplementary.
Construction: Draw a line AB parallel to ST through point R. Since AB || ST and we know that PQ || ST. Thus, AB || PQ.
In Fig. 6.31, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS. [Hint: Draw a line parallel to ST through point R.]
Let ∠SRQ = x, ∠SRB = y and ∠QRA = z
Lines ST and AB are parallel with transversal SR intersecting them. Therefore, the co-interior angles formed are supplementary.
∠RST + ∠SRB = 180°
130° + y = 180°
y = 180° - 130° = 50°
Thus, ∠SRB = y = 50°
Similarly, lines PQ and AB are parallel with transversal QR intersecting the two lines. Therefore, the co-interior angles are supplementary.
∠PQR + ∠QRA = 180°
110° + z = 180°
z = 180° - 110° = 70°
Thus, ∠QRA = z = 70°
AB is a line, RQ and RS are rays on AB. Hence,
∠QRA + ∠QRS + ∠SRB = 180°
70° + x + 50° = 180°
120° + x = 180°
x = 180° - 120° = 60°
Thus, ∠QRS = x = 60°.
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Step-by-step explanation:
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Answer:
Given: PQ || ST, ∠PQR = 110° and ∠RST = 130°
To Find: ∠QRS
Lines which are parallel to the same line are parallel to each other.
When two parallel lines are cut by a transversal, co-interior angles formed are supplementary.
Construction: Draw a line AB parallel to ST through point R. Since AB || ST and we know that PQ || ST. Thus, AB || PQ.
In Fig. 6.31, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS. [Hint: Draw a line parallel to ST through point R.]
Let ∠SRQ = x, ∠SRB = y and ∠QRA = z
Lines ST and AB are parallel with transversal SR intersecting them. Therefore, the co-interior angles formed are supplementary.
∠RST + ∠SRB = 180°
130° + y = 180°
y = 180° - 130° = 50°
Thus, ∠SRB = y = 50°
Similarly, lines PQ and AB are parallel with transversal QR intersecting the two lines. Therefore, the co-interior angles are supplementary.
∠PQR + ∠QRA = 180°
110° + z = 180°
z = 180° - 110° = 70°
Thus, ∠QRA = z = 70°
AB is a line, RQ and RS are rays on AB. Hence,
∠QRA + ∠QRS + ∠SRB = 180°
70° + x + 50° = 180°
120° + x = 180°
x = 180° - 120° = 60°
Thus, ∠QRS = x = 60°.
