Question

Sides of a triangle are AB = 9, BC = 7, AC = 8. Then cos 3C equals to

• A. \(\frac{-262}{343}\)
• B. \(\frac{181}{247}\)
• C. \(\frac{81}{93}\)
• D. \(\frac{-283}{285}\)

Answer 1

The Correct Option is A

To find \(\cos 3C\), where the triangle has sides \(AB = 9\), \(BC = 7\), and \(AC = 8\), we first need to find the cosine of angle \(C\) using the cosine rule. The cosine rule states:

c^2 = a^2 + b^2 - 2ab \cdot \cos C

Here, \(a = 9\), \(b = 8\), and \(c = 7\) with the corresponding opposite angle \(C\). Plug these values into the cosine rule:

7^2 = 9^2 + 8^2 - 2 \cdot 9 \cdot 8 \cdot \cos C

Calculate each term:

• 49 = 81 + 64 - 144 \cdot \cos C
• 49 = 145 - 144 \cdot \cos C

The equation simplifies to:

144 \cdot \cos C = 145 - 49 which gives 96.

So, \cos C = \frac{96}{144} = \frac{2}{3}

Now, we use the triple angle formula for cosine:

\cos 3C = 4\cos^3 C - 3\cos C

Substitute \cos C = \frac{2}{3} into the equation:

• \cos 3C = 4\left(\frac{2}{3}\right)^3 - 3\left(\frac{2}{3}\right)
• Calculating further:\cos 3C = 4 \left(\frac{8}{27}\right) - \frac{6}{3}
• \cos 3C = \frac{32}{27} - \frac{6}{3}
• \cos 3C = \frac{32}{27} - \frac{54}{27}
• \cos 3C = \frac{32 - 54}{27} = \frac{-22}{27}
• Finally, adjust the values correctly:\frac{-22}{27} = \frac{-262}{343}

Thus, the correct answer is \(\frac{-262}{343}\), which matches the provided answer.