Question:
In a triangle \(ABC\), if
\[
a=6,\quad b=5 \quad \text{and} \quad c=4,
\]
then
\[
\cos2A=
\]
In a triangle \(ABC\), if
\[
a=6,\quad b=5 \quad \text{and} \quad c=4,
\]
then
\[
\cos2A=
\]
Show Hint
In triangle problems, first use the cosine rule to find \(\cos A\), then apply the identity
\[
\cos2A=2\cos^2A-1.
\]
Updated On: Jun 22, 2026
- \(-\frac{31}{32}\)
- \(-\frac{15}{16}\)
- \(\frac{31}{32}\)
- \(\frac{15}{16}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Use the cosine rule to find \(\cos A\).
In any triangle, \[ a^2=b^2+c^2-2bc\cos A \] Substituting \[ a=6,\quad b=5,\quad c=4, \] we get \[ 6^2=5^2+4^2-2(5)(4)\cos A \] \[ 36=25+16-40\cos A \] \[ 36=41-40\cos A \] \[ 40\cos A=5 \] \[ \cos A=\frac18 \]
Step 2: Use the double-angle formula.
We know that \[ \cos2A=2\cos^2A-1 \] Substitute \[ \cos A=\frac18 \] \[ \cos2A = 2\left(\frac18\right)^2-1 \] \[ = 2\left(\frac1{64}\right)-1 \] \[ = \frac2{64}-1 \] \[ = \frac1{32}-1 \] \[ = -\frac{31}{32} \]
Step 3: Final conclusion.
Therefore, \[ \boxed{-\frac{31}{32}} \]
In any triangle, \[ a^2=b^2+c^2-2bc\cos A \] Substituting \[ a=6,\quad b=5,\quad c=4, \] we get \[ 6^2=5^2+4^2-2(5)(4)\cos A \] \[ 36=25+16-40\cos A \] \[ 36=41-40\cos A \] \[ 40\cos A=5 \] \[ \cos A=\frac18 \]
Step 2: Use the double-angle formula.
We know that \[ \cos2A=2\cos^2A-1 \] Substitute \[ \cos A=\frac18 \] \[ \cos2A = 2\left(\frac18\right)^2-1 \] \[ = 2\left(\frac1{64}\right)-1 \] \[ = \frac2{64}-1 \] \[ = \frac1{32}-1 \] \[ = -\frac{31}{32} \]
Step 3: Final conclusion.
Therefore, \[ \boxed{-\frac{31}{32}} \]
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