If sin(A+B) = 1 and cos (A - B)= √3/2, 0°< A+B ≤ 90° and A> B, then find the measures of angles A and B.

This question is Similar to Ex 8.2,3 - Chapter 8 Class 10

 

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Transcript

Question 25 (Choice 1) If sin (A + B) = 1 and cos (A – B)= √("3" )/2, 0°< A + B ≤ 90° and A > B, then find the measures of angles A and B.Given that sin (A + B) = 1 But we know that sin 90° = 1 Thus, sin (A + B) = sin 90° ∴ A + B = 90° cos (A – B) = √𝟑/𝟐 But we know that cos 30° = √3/2 Thus cos (A – B) = cos 30° ∴ A – B = 30° Our equations are A + B = 90° …(1) A – B = 30° …(2) Adding (1) and (2) A + B + A – B = 90° + 30° 2A = 120° A = (120°)/2 A = 60° Putting A = 60° in (1) A + B = 90° 60° + B = 90° B = 90° − 60° B = 30° Hence A = 60° , B = 30°

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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, Social Science, Physics, Chemistry, Computer Science at Teachoo.