Last updated at Dec. 16, 2024 by Teachoo
Question 5 In a Δ ABC, ∠ C = 3 ∠ B = 2 (∠ A + ∠ B). Find the three angles. Let ∠A = x & ∠B = y Given that ∠C = 3 ∠B ∠ C = 3y Also, ∠ C = 2 (∠A + ∠B) ∠C = 2 (x + y) For ∠C = 3y Put ∠A = x ∠B = y and ∠C = 3y Since ABC is a triangle, By Angle Sum property ∠A + ∠B + ∠C = 180 x + y + 3y = 180 x + 4y = 180 For ∠C = 2(x + y) Put ∠A = x ∠B = y and ∠C = 2 (x + y) Since ABC is a triangle, By Angle Sum Property ∠A + ∠B + ∠C = 180 x + y + 2 (x + y) = 180 x + y + 2x + 2y = 180 3x + 3y = 180 3(x + y) = 180 (x + y) = 180/3 x + y = 60 Hence, the equations are x + 4y = 180 …(1) x + y = 60 …(2) From equation (1) x + 4y = 180 x = 180 − 4y Put x = 180 − 4y in equation (2) x + y = 60 180 − 4y + y = 60 180 − 60 = 4y − y 120 = 3y 3y = 120 y = 120/3 y = 40 Put y = 40° in equation (1) x + 4y = 180 x + (4 × 40) = 180 x + 160 = 180 x = 20 Thus, ∠ A = x = 20° ∠ B = y = 40° ∠C = 3y = 3 × 40 = 120°
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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 and Computer Science at Teachoo