Question:
If
\[
a=2,\quad b=3,\quad c=4
\]
in a triangle \(ABC\), then
\[
\cos C=
\]
If
\[
a=2,\quad b=3,\quad c=4
\]
in a triangle \(ABC\), then
\[
\cos C=
\]
Show Hint
Use the cosine rule:
\[
c^2=a^2+b^2-2ab\cos C
\]
when all three sides of a triangle are known and an angle is required.
Updated On: Jun 22, 2026
- \(\dfrac{1}{4}\)
- \(-\dfrac{1}{4}\)
- \(\dfrac{1}{2}\)
- \(-\dfrac{1}{2}\)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Use the cosine rule.
In any triangle, \[ c^2=a^2+b^2-2ab\cos C \] Given, \[ a=2,\qquad b=3,\qquad c=4 \] Substituting these values, \[ 4^2=2^2+3^2-2(2)(3)\cos C \] \[ 16=4+9-12\cos C \] \[ 16=13-12\cos C \]
Step 2: Solve for \(\cos C\).
\[ 16-13=-12\cos C \] \[ 3=-12\cos C \] \[ \cos C=-\frac{3}{12} \] \[ \cos C=-\frac14 \]
Step 3: Final conclusion.
Hence, \[ \boxed{-\frac14} \] which corresponds to option (2).
In any triangle, \[ c^2=a^2+b^2-2ab\cos C \] Given, \[ a=2,\qquad b=3,\qquad c=4 \] Substituting these values, \[ 4^2=2^2+3^2-2(2)(3)\cos C \] \[ 16=4+9-12\cos C \] \[ 16=13-12\cos C \]
Step 2: Solve for \(\cos C\).
\[ 16-13=-12\cos C \] \[ 3=-12\cos C \] \[ \cos C=-\frac{3}{12} \] \[ \cos C=-\frac14 \]
Step 3: Final conclusion.
Hence, \[ \boxed{-\frac14} \] which corresponds to option (2).
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