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In a triangle $A B C, B C=\sqrt{39}, A C=5$ and $A B=7$. What is the measure of the angle $A ?$
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Verified Answer
The correct answer is:
$\frac{\pi}{3}$
Let $\mathrm{a}, \mathrm{b}, \mathrm{c}$ be the sides of $\Delta A B C$ and $\angle A=\theta$
$\therefore \quad a=\sqrt{39}, b=5$ and $c=7$
and $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{25+49-39}{2 \times 5 \times 7}$
$=\frac{1}{2}=\cos \frac{\pi}{3}$
$\Rightarrow \quad A=\frac{\pi}{3}$
$\therefore \quad a=\sqrt{39}, b=5$ and $c=7$
and $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{25+49-39}{2 \times 5 \times 7}$
$=\frac{1}{2}=\cos \frac{\pi}{3}$
$\Rightarrow \quad A=\frac{\pi}{3}$
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