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+1 vote
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in Mathematics by (30.6k points)

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

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1 Answer

+1 vote
by (32.0k points)

Correct answer is 39

\(\cos \mathrm{A}=\frac{\mathrm{b}^{2}+\mathrm{c}^{2}-\mathrm{a}^{2}}{2 \mathrm{bc}}\)

\(\frac{2}{3}=\frac{8^{2}+\mathrm{c}^{2}-7^{2}}{2 \times 8 \times \mathrm{c}}\)

C = 9

\(\cos \mathrm{C}=\frac{7^{2}+8^{2}-9^{2}}{2 \times 7 \times 8}=\frac{2}{7}\)

\(49 \cos 3 \mathrm{C}+42\)

\(49\left(4 \cos ^{3} \mathrm{C}-3 \cos \mathrm{C}\right)+42\)

\(49\left(4\left(\frac{2}{7}\right)^{3}-3\left(\frac{2}{7}\right)\right)+42\)

\(=\frac{32}{7}\)

\(\mathrm{m}+\mathrm{n}=32+7=39\)

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