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In a right angled triangle ABC right angled at B given 15 cosA-8sín A=0 then sin A + COS A2 cos A-sinA |
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Answer» Given that tanA= 1 3 AC= ( 3 ) 2 +1 2 = 3+1 = 4 =2 now (i)sinAcosC+cosAsinC = 2 3 × 2 3 + 2 1 × 2 1 = 4 3 + 4 1 = 4 4 =1 and (II)cosAcosC−sinAsinC = 2 1 × 2 3 − 2 1 × 2 3 = 4 3 − 4 3 =0 Step-by-step EXPLANATION: |
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