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In the triangle $\mathrm{ABC}$, if $\mathrm{a}=7, \mathrm{~b}=6$ and $\mathrm{A}=120^{\circ}$, then the approximate value of $B$ is
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Verified Answer
The correct answer is:
$47.9^{\circ}$
Given $\mathrm{a}=7, \mathrm{~b}=\mathrm{b}$ and $\mathrm{A}=120^{\circ}$.
$$
\begin{aligned}
& \frac{\sin A}{a}=\frac{\sin B}{b} \\
& \frac{\sin 120^{\circ}}{7}=\frac{\sin B}{6} \\
& \frac{\sin \left(90+30^{\circ}\right)}{7} \times 6=\sin B \\
& \sin B=\frac{6}{7} \times \cos 30^{\circ}=\frac{6}{7} \times \frac{\sqrt{3}}{2}=\frac{3 \sqrt{3}}{7}=\frac{5.196}{7}=0
\end{aligned}
$$
Required value is greater sthan to $\sin 45^{\circ}$ but less then to $\sin 60^{\circ}$.
$$
\begin{aligned}
& \frac{\sin A}{a}=\frac{\sin B}{b} \\
& \frac{\sin 120^{\circ}}{7}=\frac{\sin B}{6} \\
& \frac{\sin \left(90+30^{\circ}\right)}{7} \times 6=\sin B \\
& \sin B=\frac{6}{7} \times \cos 30^{\circ}=\frac{6}{7} \times \frac{\sqrt{3}}{2}=\frac{3 \sqrt{3}}{7}=\frac{5.196}{7}=0
\end{aligned}
$$
Required value is greater sthan to $\sin 45^{\circ}$ but less then to $\sin 60^{\circ}$.
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