Question:
In a \(\triangle ABC\), \(a = 1, b = \sqrt{3}\) and \(\angle C = \frac{\pi}{6}\). Then the measure of the third side \(c =\)
In a \(\triangle ABC\), \(a = 1, b = \sqrt{3}\) and \(\angle C = \frac{\pi}{6}\). Then the measure of the third side \(c =\)
Show Hint
Whenever two sides and the included angle of a triangle are given, the cosine rule is the fastest and most direct method to find the third side:
\[
c^2=a^2+b^2-2ab\cos C
\]
Always substitute standard trigonometric values like
\[
\cos30^\circ=\frac{\sqrt3}{2}
\]
carefully to avoid arithmetic mistakes.
Updated On: May 13, 2026
- \(4\)
- \(3\)
- \(1\)
- \(2\)
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The Correct Option is C
Solution and Explanation
Step 1: Using the cosine rule in the triangle.
For any triangle, \[ c^2 = a^2+b^2-2ab\cos C \] Substitute the given values: \[ a=1,\qquad b=\sqrt3,\qquad C=\frac{\pi}{6}=30^\circ \] Therefore, \[ c^2 = 1^2+(\sqrt3)^2-2(1)(\sqrt3)\cos30^\circ \] Now simplify step-by-step: \[ 1^2=1 \] and \[ (\sqrt3)^2=3 \] Also, \[ \cos30^\circ=\frac{\sqrt3}{2} \] Substituting: \[ c^2 = 1+3-2\sqrt3\left(\frac{\sqrt3}{2}\right) \] The \(2\) cancels: \[ c^2 = 4-\sqrt3\cdot\sqrt3 \] Since, \[ \sqrt3\cdot\sqrt3=3 \] we get: \[ c^2=4-3 \] \[ c^2=1 \] Taking positive square root because side length is positive: \[ c=1 \] Hence, \[ \boxed{c=1} \]
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