Question

In a \(\triangle ABC\), if \[ \cot\frac{A}{2}:\cot\frac{B}{2}:\cot\frac{C}{2} = 3:5:7, \] then \[ \cos A:\cos B:\cos C= \]

• A. \(2:9:12\)
• B. \(6:5:4\)
• C. \(1:2:5\)
• D. \(3:4:5\)

Hint:

In a triangle, \(\cot\frac{A}{2}=\frac{s-a}{r}\). Convert the given ratio into \((s-a):(s-b):(s-c)\), determine the sides, and then apply the cosine rule to obtain the required ratio.

Answer 1

The Correct Option is A

Concept: In a triangle, \[ \cot\frac{A}{2} = \frac{s-a}{r}, \qquad \cot\frac{B}{2} = \frac{s-b}{r}, \qquad \cot\frac{C}{2} = \frac{s-c}{r}. \] Hence, \[ (s-a):(s-b):(s-c)=3:5:7. \] Using these values, the sides can be obtained and then the cosine rule can be applied.

Step 1: Find the sides of the triangle. Let \[ s-a=3k,\qquad s-b=5k,\qquad s-c=7k. \] Adding, \[ (s-a)+(s-b)+(s-c)=15k. \] Since \[ (s-a)+(s-b)+(s-c)=s, \] we get \[ s=15k. \] Therefore, \[ a=s-3k=12k, \] \[ b=s-5k=10k, \] \[ c=s-7k=8k. \] Thus, \[ a:b:c=12:10:8=6:5:4. \]

Step 2: Find \(\cos A\). Using the cosine rule, \[ \cos A = \frac{b^2+c^2-a^2}{2bc}. \] \[ = \frac{10^2+8^2-12^2}{2(10)(8)} = \frac{100+64-144}{160} = \frac{20}{160} = \frac18. \]

Step 3: Find \(\cos B\). \[ \cos B = \frac{a^2+c^2-b^2}{2ac}. \] \[ = \frac{12^2+8^2-10^2}{2(12)(8)} = \frac{144+64-100}{192} = \frac{108}{192} = \frac{9}{16}. \]

Step 4: Find \(\cos C\). \[ \cos C = \frac{a^2+b^2-c^2}{2ab}. \] \[ = \frac{12^2+10^2-8^2}{2(12)(10)} = \frac{144+100-64}{240} = \frac{180}{240} = \frac34. \]

Step 5: Find the required ratio. \[ \cos A:\cos B:\cos C = \frac18:\frac{9}{16}:\frac34. \] Multiplying by \(16\), \[ 2:9:12. \] \[ \boxed{\cos A:\cos B:\cos C=2:9:12} \] \[ \boxed{\text{Answer = (A)}} \]