MHT CET · Maths · Properties of Triangles
In \(\triangle \mathrm{ABC}\), with usual notations, if \(\mathrm{b}=3\), \(\mathrm{c}=8, \mathrm{~m} \angle \mathrm{~A}=60^{\circ}\), then the circumradius of the triangle is __________ units.
- A \(\frac{7}{3}\)
- B \(\frac{7 \sqrt{2}}{3}\)
- C \(\frac{7}{\sqrt{3}}\)
- D \(\frac{7 \sqrt{3}}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{7}{\sqrt{3}}\)
Step-by-step Solution
Detailed explanation
By cosine rule, we get
\(\begin{aligned}
a^2 & =b^2+c^2-2 b c \cos A \\
& =9+64-48 \cos 60^{\circ} \\
& =73-48 \times \frac{1}{2} \\
& =73-24 \\
& =49 \\
a & =7
\end{aligned}\)
\(\begin{aligned}
\text {By sine rule, circumradius } & =\frac{a}{\sin \mathrm{~A}} \\
& =\frac{7}{2 \sin 60^{\circ}} \\
& =\frac{7}{2 \times \frac{\sqrt{3}}{2}} \\
& =\frac{7}{\sqrt{3}}
\end{aligned}\)
\(\begin{aligned}
a^2 & =b^2+c^2-2 b c \cos A \\
& =9+64-48 \cos 60^{\circ} \\
& =73-48 \times \frac{1}{2} \\
& =73-24 \\
& =49 \\
a & =7
\end{aligned}\)
\(\begin{aligned}
\text {By sine rule, circumradius } & =\frac{a}{\sin \mathrm{~A}} \\
& =\frac{7}{2 \sin 60^{\circ}} \\
& =\frac{7}{2 \times \frac{\sqrt{3}}{2}} \\
& =\frac{7}{\sqrt{3}}
\end{aligned}\)
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