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In a triangle $\mathrm{ABC}, \mathrm{b}=\sqrt{3} \mathrm{~cm}, \mathrm{c}=1 \mathrm{~cm}, \angle \mathrm{A}=30^{\circ}$, what is
the value of a?
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the value of a?
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Verified Answer
The correct answer is:
$1 \mathrm{~cm}$
As given: In a triangle $\mathrm{ABC}, \mathrm{AC}=\mathrm{b}=\sqrt{3} \mathrm{~cm}, \mathrm{AB}$
$=\mathrm{c}=1 \mathrm{~m}$ and $\angle \mathrm{A}=30^{\circ}$
From cosine formulae
$\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{(\sqrt{3})^{2}+1^{2}-a^{2}}{2 \sqrt{3} \cdot 1}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{3+1-a^{2}}{2 \sqrt{3}} \Rightarrow 3=4-a^{2}$
$\Rightarrow \mathrm{a}^{2}=4-3=1 \Rightarrow \mathrm{a}=1 \mathrm{~cm}$
$=\mathrm{c}=1 \mathrm{~m}$ and $\angle \mathrm{A}=30^{\circ}$
From cosine formulae
$\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{(\sqrt{3})^{2}+1^{2}-a^{2}}{2 \sqrt{3} \cdot 1}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{3+1-a^{2}}{2 \sqrt{3}} \Rightarrow 3=4-a^{2}$
$\Rightarrow \mathrm{a}^{2}=4-3=1 \Rightarrow \mathrm{a}=1 \mathrm{~cm}$
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