In each of the figures given below, AB || CD. Find the unknown angles, giving reasons.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)

Question

In each of the figures given below, AB || CD. Find the unknown angles, giving reasons. (i) (ii) (iii) (iv) (v) (vi) (vii) (viii)

KnowledgeBoat · Accepted Answer

(i) Given: AB || CD ∠ECD = 75° ∠AEC = ∠ECD [Alternate angles] ⇒ y° = ∠ECD ⇒ y° = 75° [Substituting the value of ∠ECD] Since, AB is a straight line: ∠BEC + ∠AEC = 180° [Linear pair] ⇒ x° + y° = 180° ⇒ x° + 75° = 180° [Substituting the value of y] ⇒ x° = 180° - 75° ⇒ x° = 105° x° = 105° and y° = 75° (ii) Given: AB || CD ∠EAB = 130° ∠BCD = 70° Since, EF is a straight line: ∠EAB + ∠BAC = 180° [Linear pair] ⇒ 130° + ∠BAC = 180° [Substituting the value of ∠EAB] ⇒ ∠BAC = 180° - 130° ⇒ ∠BAC = 50° ∠DCF = ∠BAC [Corresponding angles] ⇒ x° = ∠BAC ⇒ x° = 50° [Substituting the value of ∠BAC] Since EF is a straight line: ∠BCA + ∠BCF = 180° [Linear pair] ⇒ y° + (x° + 70°) = 180° ⇒ y° + 50° + 70° = 180° [Substituting the value of x] ⇒ y° + 120° = 180° ⇒ y° = 180° - 120° ⇒ y° = 60° In △ABC, we know that the sum of interior angles is 180°. ∴ ∠BAC + ∠ABC + ∠BCA = 180° ⇒ 50° + z° + y° = 180° ⇒ 50° + z° + 60° = 180° [Substituting the value of y] ⇒ z° + 110° = 180° ⇒ z° = 180° - 110° ⇒ z° = 70° x° = 50°, y° = 60° and z° = 70° (iii) Given: ∠CQP = 120° ∠DRP = 115° Since, CD is a straight line: ∠CQP + ∠PQR = 180° [Linear pair] ⇒ 120° + ∠PQR = 180° [Substituting the value of ∠CQP] ⇒ ∠PQR = 180° - 120° ⇒ ∠PQR = 60° AB || CD and PQ is a transversal: ∠APQ = ∠PQR [Alternate angles] ⇒ x° = ∠PQR ⇒ x° = 60° [Substituting the value of ∠PQR] Since, CD is a straight line: ∠PRD + ∠PRQ = 180° [Linear pair] ⇒ 115° + ∠PRQ = 180° [Substituting the value of ∠PRD] ⇒ ∠PRQ = 180° - 115° ⇒ ∠PRQ = 65° AB || CD and PR is a transversal: ∠BPR = ∠PRQ [Alternate angles] ⇒ z° = ∠PRQ ⇒ z° = 65° [Substituting the value of ∠PRQ] In △PQR, we know that the sum of interior angles is 180°. ∴ ∠PQR + ∠QPR + ∠PRQ = 180° ⇒ 60° + y° + 65° = 180° ⇒ y° + 125° = 180° ⇒ y° = 180° - 125° ⇒ y° = 55° x° = 60°, y° = 55° and z° = 65° (iv) Given: ∠DCE = 100° ∠BCA = 30° Since, AE is a straight line: ∠BCA + ∠BCD + ∠DCE = 180° ⇒ 30° + z° + 100° = 180° ⇒ z° + 130° = 180° ⇒ z° = 180° - 130° ⇒ z° = 50° AB || CD and BC is a transversal: ∠ABC = ∠BCD [Alternate angles] ⇒ x° = z° ⇒ x° = 50° In △ABC, we know that the sum of interior angles is 180°. ∴ ∠ABC + ∠BAC + ∠BCA = 180° ⇒ x° + y° + 30° = 180° ⇒ 50° + y° + 30° = 180° ⇒ y° + 80° = 180° ⇒ y° = 180° - 80° ⇒ y° = 100° x° = 50°, y° = 100° and z° = 50° (v) From figure, ∠BAD = 35° ∠ECD = 75° AB || CD and AD is a transversal: ∠ADC = ∠BAD [Alternate angles] ∠ADC = 35° ∴ ∠CDE = 35° [Both are same angles i.e., ∠D] In △CDE, we know that the sum of interior angles is 180°. ∴ ∠ECD + ∠CDE + ∠DEC = 180° ⇒ 75° + 35° + x° = 180° ⇒ x° + 110° = 180° ⇒ x° = 180° - 110° x° = 70° (vi) Through M, draw a line EMF such that EF || AB || CD. This line splits the angle x into two parts: Let ∠EMB = z° and ∠EMD = y° Now, EF || CD and MD is a transversal. ∴ The sum of co-interior angles is 180°. ∴ ∠CDM + ∠DME = 180° ⇒ 130° + y° = 180° ⇒ y° = 180° - 130° ⇒ y° = 50° Again, Now, EF || AB and MB is a transversal. ∴ The sum of co-interior angles is 180°. ∴ ∠ABM + ∠BME = 180° ⇒ 150° + z° = 180° ⇒ z° = 180° - 150° ⇒ z° = 30° x° = y° + z° ⇒ x° = 50° + 30° x° = 80° (vii) Through M, draw a line EMF such that EF || AB || CD. This line splits the angle at M into two parts: Let ∠EMB = y° and ∠EMD = (90 - y)° Now, EF || CD and MD is a transversal. ∴ The sum of co-interior angles is 180°. ∴ ∠CDM + ∠DME = 180° ⇒ 140° + (90 - y)° = 180° ⇒ 140° + 90° - y° = 180° ⇒ 230° - y° = 180° ⇒ y° = 230° - 180° ⇒ y° = 50° Again, EF || AB and MB is a transversal. ∴ The sum of co-interior angles is 180°. ∴ ∠ABM + ∠BME = 180° ⇒ x° + y° = 180° ⇒ x° + 50° = 180° ⇒ x° = 180° - 50° x° = 130° (viii) Through M, draw a line EMF such that EF || AB || CD. This line splits the angle at M into two parts: Let ∠AME = y° and ∠CME = z° Now, EF || AB and AM is a transversal. ∠AME = ∠BAM [Alternate angles] ∴ y° = 40° Again, EF || CD and CM is a transversal. ∠CME = ∠MCD [Alternate angles] ∴ z° = 50° x° = y° + z° ⇒ x° = 40° + 50° x° = 90°

Mathematics

In each of the figures given below, AB || CD. Find the unknown angles, giving reasons.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)

Lines & Angles

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