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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

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 R.) It is given that PQ || ST, We draw a line XY || ST , So, XY || PQ , i.e. PQ || ST || XY Since PQ || XY & QR is the transversal So, ∠PQR + ∠QRX = 180° 110° + ∠ QRX = 180° ∠ QRX = 180° – 110° ∠ QRX = 70° Also, ST || XY & SR is the transversal ∠SRY + ∠RST = 180° 130° + ∠ SRY= 180° ∠ SRY= 180° – 130° ∠ SRY = 50° Since XY is a line ∠QRX + ∠QRS + ∠SRY = 180° 70° + ∠QRS + 50° = 180° 120° + ∠QRS = 180° ∠ QRS = 180° – 120° ∠ QRS = 60°

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.