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Question

Mathematics Question on Triangles

In a triangle ABCABC, BC=7BC = 7, AC=8AC = 8, AB=αNAB = \alpha \in \mathbb{N} and cosA=23\cos A = \frac{2}{3}. If 49cos(3C)+42=mn,49 \cos(3C) + 42 = \frac{m}{n}, where gcd(m,n)=1\gcd(m, n) = 1, then m+nm + n is equal to ________.

Answer

Using the cosine rule for cosA\cos A:
cosA=b2+c2a22bc\cos A = \frac{b^2 + c^2 - a^2}{2bc}
Substitute b=8b = 8, c=7c = 7, and cosA=23\cos A = \frac{2}{3}:
23=82+72a22×8×7\frac{2}{3} = \frac{8^2 + 7^2 - a^2}{2 \times 8 \times 7}
    a2=9\implies a^2 = 9
    a=3\implies a = 3
Now, calculate cosC\cos C using the cosine rule:
cosC=72+82922×7×8=27\cos C = \frac{7^2 + 8^2 - 9^2}{2 \times 7 \times 8} = \frac{2}{7}
Then, for cos(3C)\cos(3C), we use the triple angle formula:
49cos(3C)+42=49(4cos3C3cosC)+4249 \cos(3C) + 42 = 49 \left( 4 \cos^3 C - 3 \cos C \right) + 42
Substituting cosC=27\cos C = \frac{2}{7}:
=49(4(27)3327)+42= 49 \left( 4 \left( \frac{2}{7} \right)^3 - 3 \cdot \frac{2}{7} \right) + 42
=49(3234367)+42= 49 \left( \frac{32}{343} - \frac{6}{7} \right) + 42
=327+42= \frac{32}{7} + 42
Thus, m=32m = 32 and n=7n = 7, so:
m+n=32+7=39m + n = 32 + 7 = 39