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In a $\triangle A B C, a=2 \mathrm{~cm}, b=3 \mathrm{~cm}, c=4 \mathrm{~cm}$ then $\cos A$ equals to -
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Verified Answer
The correct answer is:
$\frac{7}{8}$
By the cosine rule,
$\cos A=\frac{b^2+c^2-a^2}{2 b c}$
$\cos A=\frac{3^2+4^2-2^2}{2(3)(4)}$
$\cos A=\frac{21}{24}$
$\cos A=\frac{7}{8}$
$\cos A=\frac{b^2+c^2-a^2}{2 b c}$
$\cos A=\frac{3^2+4^2-2^2}{2(3)(4)}$
$\cos A=\frac{21}{24}$
$\cos A=\frac{7}{8}$
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