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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