(A) Given in $\triangle ABC$: $a=6$,$b=5$,$c=4$. Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$. Substituting the values: $\cos A =

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(A) Given in $\triangle ABC$: $a=6$,$b=5$,$c=4$. Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$. Substituting the values: $\cos A = \frac{5^2 + 4^2 - 6^2}{2 \times 5 \times 4} = \frac{25 + 16 - 36}{40} = \frac{5}{40} = \frac{1}{8}$. Now,using the formula $\cos 2A = 2\cos^2 A - 1$: $\cos 2A = 2 \left(\frac{1}{8}\right)^2 - 1 = 2 \left(\frac{1}{64}\right) - 1 = \frac{1}{32} - 1 = \frac{1 - 32}{32} = -\frac{31}{32}$.

Class 11 Mathematics Trigonometrical Equations Relation between sides and angles, Solutions of triangles

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In a $\triangle ABC$,if $a=6$,$b=5$,and $c=4$,then find the value of $\cos 2A$.

AP EAMCET 2022 All AP EAMCET

A
$-\frac{31}{32}$
B
$-\frac{15}{16}$
C
$\frac{31}{32}$
D
$\frac{15}{16}$

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Relation between sides and angles, Solutions of triangles

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In a $\triangle ABC$,if $a=6$,$b=5$,and $c=4$,then find the value of $\cos 2A$.

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