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In a $\triangle A B C$, $a+b=3(1+\sqrt{3}) \mathrm{cm}$ and $a-b=3(1-\sqrt{3}) \mathrm{cm}$.
If angle $A$ is $30^{\circ}$, then what is the angle $B$ ?
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If angle $A$ is $30^{\circ}$, then what is the angle $B$ ?
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Verified Answer
The correct answer is:
$60^{\circ}$
Given, $a+b=3(1+\sqrt{3}) \ldots \ldots \ldots$. (1)
and $a-b=3(1-\sqrt{3}) \ldots \ldots \ldots .(2)$
By adding (1) and (2) we get
$(a+b)+(a-b)=3+3 \sqrt{3}+3-3 \sqrt{3}$
$\Rightarrow 2 a=6 \Rightarrow a=3$
$\therefore b=3-3+3 \sqrt{3}=3 \sqrt{3}$
By using sine rule, $\Rightarrow \frac{a}{\sin A}=\frac{b}{\sin B}$
Given $\angle A=30 \quad \Rightarrow \frac{3}{\sin 30^{\circ}}=\frac{3 \sqrt{3}}{\sin B}$
$\Rightarrow \sin B=\sqrt{3} \times \frac{1}{2} \Rightarrow \sin B=\sin 60^{\circ}$
$\Rightarrow B=60^{\circ}$
and $a-b=3(1-\sqrt{3}) \ldots \ldots \ldots .(2)$
By adding (1) and (2) we get
$(a+b)+(a-b)=3+3 \sqrt{3}+3-3 \sqrt{3}$
$\Rightarrow 2 a=6 \Rightarrow a=3$
$\therefore b=3-3+3 \sqrt{3}=3 \sqrt{3}$
By using sine rule, $\Rightarrow \frac{a}{\sin A}=\frac{b}{\sin B}$
Given $\angle A=30 \quad \Rightarrow \frac{3}{\sin 30^{\circ}}=\frac{3 \sqrt{3}}{\sin B}$
$\Rightarrow \sin B=\sqrt{3} \times \frac{1}{2} \Rightarrow \sin B=\sin 60^{\circ}$
$\Rightarrow B=60^{\circ}$
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