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If in ABC\triangle ABC A+B=90\angle A + \angle B = 90^\circ and sinB=45\sin B = \frac{4}{5}, then find the value of cosA\cos A.

Solution

Correct Option: 3

Given A+B=90\angle A + \angle B = 90^\circ, so A=90B\angle A = 90^\circ - \angle B.

Therefore cosA=cos(90B)=sinB=45\cos A = \cos(90^\circ - B) = \sin B = \frac{4}{5}.

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