Question

In a triangle \(ABC\), if \[ a=2,\qquad b=4,\qquad \cos C=-\frac{5}{16}, \] then the circumradius \(R\) is

• A. \(\dfrac{30}{\sqrt{231}}\)
• B. \(\dfrac{40}{\sqrt{231}}\)
• C. \(\dfrac{20}{\sqrt{231}}\)
• D. \(\dfrac{10}{\sqrt{231}}\) \bigskip

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.

Answer 1

The Correct Option is B

Concept: Use the Law of Cosines to find side \(c\), then the formula for circumradius \(R = \frac{abc}{4\Delta}\) where area \(\Delta = \frac{1}{2}ab\sin C\).

Step 1: Find side \(c\) using Law of Cosines. Given \(a=2, b=4, \cos C = -\frac{5}{16}\). \[ c^2 = a^2 + b^2 - 2ab\cos C = 4 + 16 - 2(2)(4)\left(-\frac{5}{16}\right) = 20 + 5 = 25 \implies c = 5. \]

Step 2: Find \(\sin C\) and area \(\Delta\). \[ \sin C = \sqrt{1 - \cos^2 C} = \sqrt{1 - \frac{25}{256}} = \sqrt{\frac{231}{256}} = \frac{\sqrt{231}}{16}. \] \[ \Delta = \frac{1}{2}ab\sin C = \frac{1}{2}(2)(4)\cdot\frac{\sqrt{231}}{16} = \frac{\sqrt{231}}{4}. \]

Step 3: Compute circumradius \(R\). Using \(R = \frac{abc}{4\Delta}\): \[ R = \frac{(2)(4)(5)}{4 \cdot \frac{\sqrt{231}}{4}} = \frac{40}{\sqrt{231}}. \]

Step 4: Write the final answer. \[ \boxed{\frac{40}{\sqrt{231}}} \]