If pq st pqr 110 and rst 130 find qrs
WebIn Fig. 6, if PQ ST, PQR = 110° and RST = 130°, find QRS. [Hint : Draw a line parallel to ST through point R.] Solution: First, construct a line XY parallel to PQ. We know that the angles on the same side of transversal is equal to 180°. So, PQR+ QRX = 180° Or, QRX = 180°-110° ∴ QRX = 70° Angles. In Fig. 6, PQ and RS are two mirrors ... Web10 okt. 2024 · $PQRS$ is a trapezium in which $PQ SR$ and $angle P=130^{circ}, angle Q=110^{circ}$. Then find $angle R$ and $angle S$.
If pq st pqr 110 and rst 130 find qrs
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Web4. In Fig. 6.31, if PQ ST, PQR = 110° and RST = 130°, find QRS. [Hint : Draw a line parallel to ST through point R.] Solution: First, construct a line XY parallel to PQ. We know that the angles on the same side of transversal is equal to 180°. So, PQR+ QRX = 180° Or, QRX = 180°-110° ∴ QRX = 70° WebQ4 In Fig. 6.31, if PQ ST, PQR = 110° and RST = 130°, find QRS. [ Hint : Draw a line parallel to ST through point R.] Answer: Draw a line EF parallel to the ST through R.
WebPQ is a transversal. So, APQ = PQR= (alternate interior angles) Again, PR is a transversal. So, y + = (Alternate interior angles) Q6 In Fig. 6.33, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes WebClick here👆to get an answer to your question ️ In Fig. \( 6.31 , \) if PQ \( \left\ \) ST, \( \angle P Q R = 110 ^ { \circ } \) and \right \( \angle R S T = 130 ...
Web17 okt. 2024 · SOLUTION : Given :PQ ST, ∠PQR = 110° and ∠RST = 130°. Construction:A line XY parallel to PQ and ST is drawn. ∠PQR + ∠QRX = 180° (Angles … WebExercise 6.1 (Page 96) 1. In the following figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE. Ans. Since AB is a straight line, Thus, ∠BOE = 30° and reflex ∠COE = 250. 2. In the …
Web8 sep. 2024 · In Fig. 3.31, If PQ ST, ∠PQR =110°, and ∠RST = 130°, find ∠QRS. (Hint: Draw a line parallel to ST through point R). Answer: Data: PQ ST and ∠PQR= 110° and ∠RST= 130°. To Prove: ∠QRS =? Construction: Draw ST ...
Web16 jul. 2024 · KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2. July 16, 2024 / By Prasanna. Students can Download Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 9 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, … mxnitro ブラウザWeb11 sep. 2024 · 4. In Fig. 6.31, if PQ ST, PQR = 110° and RST = 130°, find QRS. [Hint : Draw a line parallel to ST through point R.] Solution: First, construct a line XY parallel to PQ. We know that the angles on the same side of transversal is equal to 180°. So, PQR+QRX = 180° Or,QRX = 180°-110° ∴ QRX = 70° Similarly, RST +SRY = 180° Or, SRY = 180 ... mxn jpy ピボットWeb21 mrt. 2024 · In the given figure $PQ$ is parallel to $ST$, $\angle PQR = {110^ \circ }{\text{ and }}\angle RST = {130^ \circ }$. Find $\angle QRS$. mxnl2j マウスWeb7 sep. 2024 · KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2. kseebsol. September 7, 2024. Class 9. KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Exercise 3.2. mxn/jpy スワップポイントWebIn Fig. below, if PQ ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠ QRS. [Hint : Draw a line parallel to ST through point R.] mxms マットWeb28 mrt. 2024 · Ex 6.2, 4 In the given figure, if PQ ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS. (Hint: Draw a line parallel to ST through point … mxone ダウンロードWeb11 apr. 2024 · EXERCISE 15A Find the volume, the lateral surface area and the total the cuboid whose dimensions are: (i) length = 12 cm, breadth = 8 cm and height = 4.5 cm (ii) length = 26 m, breadth = 14 m and height = 6.5 m iii) length = 15 m, breadth = 6 m and height = 5 dm iv) length = 24 m, breadth = 25 cm and height = 6 m A matchbox … mxmlファイル