Question : In the triangle $\mathrm{ABC}, \mathrm{AB}=12 \mathrm{~cm},\mathrm{AC}=10 \mathrm{~cm}$, and $\angle \mathrm{BAC}=60^{\circ}$. What is the value of the length of the side $\mathrm{BC}$?
Option 1: 10 cm
Option 2: 7.13 cm
Option 3: 13.20 cm
Option 4: 11.13 cm
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Correct Answer: 11.13 cm
Solution : Given, $\mathrm{AB}=12 \mathrm{~cm},\mathrm{AC}=10 \mathrm{~cm}$, and $\angle \mathrm{BAC}=60^{\circ}$ According to cosine law, $\mathrm{BC}^2=\mathrm{AB}^2+\mathrm{AC}^2-2(\mathrm{AB})(\mathrm{AC})\cos 60^\circ$ ⇒ $\mathrm{BC}^2=12^2+10^2-2\times12\times10\times0.5$ $\therefore \mathrm{BC}=11.13\mathrm{~cm}$ Hence, the correct answer is 11.13 cm.
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Question : $ABC$ is a triangle and $D$ is a point on the side $BC$. If $BC = 16\mathrm{~cm}$, $BD = 11 \mathrm{~cm}$ and $\angle ADC = \angle BAC$, then the length of $AC$ is equal to:
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