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

Solution

✅ Correct Option: 3

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

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

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