ANGLES so ∠BEP = ∠DFE and ∠DFE and ∠DFQ is a straight line angle.Calculation∠DFE = x⇒ y + x = 1800 ⇒ ∠DFQ + ∠DFE = 1800⇒ 3x + x = 1800⇒ 4X = 1800 ⇒ x = 1800/4 = 450Now,⇒ y + 450 = 1800⇒ y = 1800 - 450 ⇒ y = 1350∴ y = 1350

"> ANGLES so ∠BEP = ∠DFE and ∠DFE and ∠DFQ is a straight line angle.Calculation∠DFE = x⇒ y + x = 1800 ⇒ ∠DFQ + ∠DFE = 1800⇒ 3x + x = 1800⇒ 4X = 1800 ⇒ x = 1800/4 = 450Now,⇒ y + 450 = 1800⇒ y = 1800 - 450 ⇒ y = 1350∴ y = 1350

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AB and CD are two parallel lines. The transversal PQ curs AB at E and CD at F. if \(\angle AEP=y,\angle PEB=x, \angle DFQ = 3x,\)then the value of y is:

SRMJEEE Analytical Geometry in SRMJEEE 9 months ago

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Given∠APE = y, ∠PEB = x, ∠DFQ = 3xConceptAs PQ is transversal line AB is a straight line angle and,∠BEP and ∠DFE is corresponding ANGLES so ∠BEP = ∠DFE and ∠DFE and ∠DFQ is a straight line angle.Calculation∠DFE = x⇒ y + x = 1800 ⇒ ∠DFQ + ∠DFE = 1800⇒ 3x + x = 1800⇒ 4X = 1800 ⇒ x = 1800/4 = 450Now,⇒ y + 450 = 1800⇒ y = 1800 - 450 ⇒ y = 1350∴ y = 1350

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Posted on 01 Nov 2024, this text provides information on SRMJEEE related to Analytical Geometry in SRMJEEE. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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