Trigonometry & Inverse Trigonometry
Cosine Rule in Triangles
Grade 11
Question:
<p>Given, <span style='display:inline-block;border-bottom:1px solid;'>$\frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13}$</span> for a triangle ABC with <span style='display:inline-block;border-bottom:1px solid;'>$\frac{\cos A}{\alpha} = \frac{\cos B}{\beta} = \frac{\cos C}{\gamma}$</span>, then the ordered triad $(\alpha, \beta, \gamma)$ has a value</p>
<p>(a) (19, 7, 25)</p>
<p>(b) (3, 4, 5)</p>
<p>(c) (5, 12, 13)</p>
<p>(d) (7, 19, 25)</p>
Step-by-Step Solution
Key Concept: Express the given ratios as multiples of a constant, solve the system to find side lengths, then apply the Cosine Rule to find the cosines and establish the ratios of the parameters.
<p>Let $\frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13} = k$ (say)</p><p>Then $b+c = 11k$, $c+a = 12k$, $a+b = 13k$</p><p>Adding all three: $2(a+b+c) = 36k$, so $a+b+c = 18k$</p><p>Therefore: $a = 18k - 11k = 7k$, $b = 18k - 12k = 6k$, $c = 18k - 13k = 5k$</p><p>Using the Cosine Rule:</p><p>$\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{36k^2+25k^2-49k^2}{2(6k)(5k)} = \frac{12k^2}{60k^2} = \frac{1}{5}$</p><p>$\cos B = \frac{c^2+a^2-b^2}{2ca} = \frac{25k^2+49k^2-36k^2}{2(5k)(7k)} = \frac{38k^2}{70k^2} = \frac{19}{35}$</p><p>$\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{49k^2+36k^2-25k^2}{2(7k)(6k)} = \frac{60k^2}{84k^2} = \frac{5}{7}$</p><p>From $\frac{\cos A}{\alpha} = \frac{\cos B}{\beta} = \frac{\cos C}{\gamma}$:</p><p>$\frac{1/5}{\alpha} = \frac{19/35}{\beta} = \frac{5/7}{\gamma}$</p><p>This gives $(\alpha, \beta, \gamma) = (7, 19, 25)$</p><p>∴ Answer is (d)</p>
Correct Answer: D