Question 62

If in a triangle ABC, Sin A = Cos B then the value of Cos C is

Solution

If in a triangle ABC, Sin A = Cos B

=> $$sin(A)=sin(90^\circ-B)$$

=> $$A=90^\circ-B$$

=> $$A+B=90^\circ$$ -----------(i)

Also, in $$\triangle$$ ABC, => $$A+B+C=180^\circ$$

Substituting value from equation (i),

=> $$C=180-90=90^\circ$$

$$\therefore$$ $$cos(C)=cos(90^\circ)=0$$

=> Ans - (B)

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