In a triangle ABC,if a=6,b=5,and c=4,then fin | vedclass

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(A) Given in $	riangle 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 =

Vedclass · Accepted Answer

$-\frac{31}{32}$

Vedclass · Answer

(A) Given in $	riangle 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: $\cos A = \frac{5^2 + 4^2 - 6^2}{2 	imes 5 	imes 4} = \frac{25 + 16 - 36}{40} = \frac{5}{40} = \frac{1}{8}$. Now,using the formula $\cos 2A = 2\cos^2 A - 1$: $\cos 2A = 2 \left(\frac{1}{8}ight)^2 - 1 = 2 \left(\frac{1}{64}ight) - 1 = \frac{1}{32} - 1 = \frac{1 - 32}{32} = -\frac{31}{32}$.

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

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