If 3sinA+4cosA=5, find sinA
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3sinA+4cosA=5
divide both side by cosA
3sinA/cosA +4cosA/cosA=5/cosA
3tanA+4=5secA
take square both side
9tan^2A+16+24tanA=25sec^2A
we know
sec^2A=1+tan^2A
so,
9tan^2A+16+24tanA=25+25tan^2A
16tan^2A-24tanA+9=0
(4tanA-3)^2=0
tanA=3/4
so,
h=√(3^2+4^2)=5
simA=p/h=3/5
divide both side by cosA
3sinA/cosA +4cosA/cosA=5/cosA
3tanA+4=5secA
take square both side
9tan^2A+16+24tanA=25sec^2A
we know
sec^2A=1+tan^2A
so,
9tan^2A+16+24tanA=25+25tan^2A
16tan^2A-24tanA+9=0
(4tanA-3)^2=0
tanA=3/4
so,
h=√(3^2+4^2)=5
simA=p/h=3/5
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