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
- $3/5$
- $4/5$
- $3/4$
- $4/3$
Solution
Opp to $C$ is $AB=3$, hyp $=5$.
Asked in: MH-SSC-9
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