Question : In $\triangle ABC$, $\angle B = 90^{\circ}$, AB = 8 cm and BC = 15 cm, then $\sin C$?
Option 1: $\frac{15}{17}$
Option 2: $\frac{8}{17}$
Option 3: $\frac{15}{8}$
Option 4: $\frac{8}{15}$
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Correct Answer: $\frac{8}{17}$
Solution :
By Pythagoras theorem,
AC = $\sqrt{8^2 + 15^2}$
= $\sqrt{64 + 225}$
= $\sqrt{289}$
= 17 cm
⇒ $\sin$ C = $\frac{8}{17}$
Hence, the correct answer is $\frac{8}{17}$.
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