In a triangle ABC,BC=7,AC=8,AB=alpha in N and | vedclass

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(B) Using the Law of Cosines: $\cos A = \frac{b^2+c^2-a^2}{2bc}$.Given $a=7$,$b=8$,$c=\alpha$,and $\cos A = \frac{2}{3}$:$\frac{2}{3} = \frac{8^2+\alp

Vedclass · Accepted Answer

$39$

Vedclass · Answer

(B) Using the Law of Cosines: $\cos A = \frac{b^2+c^2-a^2}{2bc}$.Given $a=7$,$b=8$,$c=\alpha$,and $\cos A = \frac{2}{3}$:$\frac{2}{3} = \frac{8^2+\alpha^2-7^2}{2 	imes 8 	imes \alpha} = \frac{64+\alpha^2-49}{16\alpha} = \frac{15+\alpha^2}{16\alpha}$.$32\alpha = 45 + 3\alpha^2 \implies 3\alpha^2 - 32\alpha + 45 = 0$.$(3\alpha - 5)(\alpha - 9) = 0$. Since $\alpha \in N$,we have $\alpha = 9$.Now,find $\cos C$ using the Law of Cosines: $\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{7^2+8^2-9^2}{2 	imes 7 	imes 8} = \frac{49+64-81}{112} = \frac{32}{112} = \frac{2}{7}$.We need to evaluate $49 \cos(3C) + 42$.Using $\cos(3C) = 4\cos^3 C - 3\cos C$:$49(4(\frac{2}{7})^3 - 3(\frac{2}{7})) + 42 = 49(4 	imes \frac{8}{343} - \frac{6}{7}) + 42 = 49(\frac{32}{343} - \frac{6}{7}) + 42 = 49(\frac{32 - 294}{343}) + 42 = \frac{32 - 294}{7} + 42 = \frac{-262}{7} + \frac{294}{7} = \frac{32}{7}$.Thus,$\frac{m}{n} = \frac{32}{7}$,so $m=32$ and $n=7$. Since $\operatorname{gcd}(32, 7)=1$,$m+n = 32+7 = 39$.

In a triangle $ABC$,$BC=7$,$AC=8$,$AB=\alpha \in N$ and $\cos A=\frac{2}{3}$. If $49 \cos (3C)+42=\frac{m}{n}$,where $\operatorname{gcd}(m, n)=1$,then $m+n$ is equal to..........

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