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asked Jul 5, 2016 in Trigonometry Answers by Srijan

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3sinA=5-4cosA; square both sides: 9sin^2(A)=25-40cosA+16cos^2(A);

9-9cos^2(A)=25-40cosA+16cos^2(A); 25cos^2(A)-40cosA+16=0=(5cosA-4)^2.

cosA=4/5 so sinA=3/5 (√(1-16/25) and sinA+cosA=7/5=1.4.

CHECK: 3sinA+4cosA=9/5+16/5=25/5=5. OK
answered Jul 6, 2016 by Rod Top Rated User (486,900 points)
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