Question

in a right angled triangle ABC, angle A is acute, angle B =90° and tanA=3/5 sinA​

Answer 1

ANSWER⤵️

tan A=(2tanA/2)÷(1-tan^2A/2)=3/4, or 8tanA/2=3-3tan^2A/2,or 3tan^2A/2+8tanA/2-3=0 or, 3tan^2A/2+9tanA/2-tanA/2-3=0, or 3tanA/2(tanA/2+3)-1(tanA/2+3)=0, or (3tanA/2-1)(tanA/2+3)=0 so tanA/2=1/3,-3. Now sinA=2tanA/2÷(1+tan^2A/2) & cosA=(1-tan^2A/2)÷(1+tan^2A/2) when tanA/2=1/3 then sinA=2×1/3÷(1+1/9)=2/3×9/10=3/5. So cosA=(1-1/9)÷(1+1/9)=8/9×9/10=8/10=4/5. When tanA/2=-3 then sinA=2(-3)/(1+9)=-6/10=(-3/5) & cosA=(1–9)/(1+9)=-8/10=(-4/5)

Ans: sinA=3/5 & -3/5 & cosA=4/5,-4/5

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