Question:
In a triangle \(ABC\), if
\[
a=2,\qquad b=4,\qquad \cos C=-\frac{5}{16},
\]
then the circumradius \(R\) is
In a triangle \(ABC\), if
\[
a=2,\qquad b=4,\qquad \cos C=-\frac{5}{16},
\]
then the circumradius \(R\) is
Show Hint
When two sides and the included angle are given:
\[
c^2=a^2+b^2-2ab\cos C
\]
first find the third side. Then use
\[
c=2R\sin C
\]
to obtain the circumradius directly.
Updated On: Jul 9, 2026
- \(\dfrac{30}{\sqrt{231}}\)
- \(\dfrac{40}{\sqrt{231}}\)
- \(\dfrac{20}{\sqrt{231}}\)
- \(\dfrac{10}{\sqrt{231}}\) \bigskip
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
For a triangle,
\[
c^2=a^2+b^2-2ab\cos C
\]
(Law of Cosines)
and
\[
c=2R\sin C
\]
(Extended Law of Sines).
Using these two formulas, we can find the circumradius \(R\).
Step 1: Find the side \(c\) using the Law of Cosines. Given \[ a=2,\qquad b=4,\qquad \cos C=-\frac{5}{16}. \] Therefore, \[ c^2 = a^2+b^2-2ab\cos C. \] \[ = 2^2+4^2-2(2)(4)\left(-\frac{5}{16}\right). \] \[ = 4+16+5. \] \[ =25. \] Hence, \[ c=5. \]
Step 2: Find \(\sin C\). Using \[ \sin^2C+\cos^2C=1, \] we get \[ \sin C = \sqrt{1-\left(-\frac{5}{16}\right)^2}. \] \[ = \sqrt{1-\frac{25}{256}}. \] \[ = \sqrt{\frac{231}{256}}. \] \[ = \frac{\sqrt{231}}{16}. \] Since \(C\) is an angle of a triangle, \[ \sin C>0. \] Thus, \[ \sin C=\frac{\sqrt{231}}{16}. \]
Step 3: Use the Extended Law of Sines. \[ c=2R\sin C. \] Substituting \(c=5\), \[ 5 = 2R\left(\frac{\sqrt{231}}{16}\right). \] \[ 5 = R\left(\frac{\sqrt{231}}{8}\right). \] Therefore, \[ R = \frac{40}{\sqrt{231}}. \]
Step 4: Write the final answer. \[ \boxed{\frac{40}{\sqrt{231}}} \]
Step 1: Find the side \(c\) using the Law of Cosines. Given \[ a=2,\qquad b=4,\qquad \cos C=-\frac{5}{16}. \] Therefore, \[ c^2 = a^2+b^2-2ab\cos C. \] \[ = 2^2+4^2-2(2)(4)\left(-\frac{5}{16}\right). \] \[ = 4+16+5. \] \[ =25. \] Hence, \[ c=5. \]
Step 2: Find \(\sin C\). Using \[ \sin^2C+\cos^2C=1, \] we get \[ \sin C = \sqrt{1-\left(-\frac{5}{16}\right)^2}. \] \[ = \sqrt{1-\frac{25}{256}}. \] \[ = \sqrt{\frac{231}{256}}. \] \[ = \frac{\sqrt{231}}{16}. \] Since \(C\) is an angle of a triangle, \[ \sin C>0. \] Thus, \[ \sin C=\frac{\sqrt{231}}{16}. \]
Step 3: Use the Extended Law of Sines. \[ c=2R\sin C. \] Substituting \(c=5\), \[ 5 = 2R\left(\frac{\sqrt{231}}{16}\right). \] \[ 5 = R\left(\frac{\sqrt{231}}{8}\right). \] Therefore, \[ R = \frac{40}{\sqrt{231}}. \]
Step 4: Write the final answer. \[ \boxed{\frac{40}{\sqrt{231}}} \]
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