In right $\triangle ABC$, $\angle B=90°$, $AB=3, BC=4, AC=5$. Then $\sin C$ equals

In right $\triangle ABC$, $\angle B=90°$, $AB=3, BC=4, AC=5$. Then $\sin C$ equals
  1. $3/5$
  2. $4/5$
  3. $3/4$
  4. $4/3$

Solution

Opp to $C$ is $AB=3$, hyp $=5$.

Asked in: MH-SSC-9

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