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