Multiple choice

lf in $\Delta$ $ABC$ $\displaystyle \cot\frac{A}{2}$ : $\displaystyle \cot\frac{B}{2}$ : $\displaystyle \cot\frac{C}{2}=1:4:15$, then the greatest angle of triangle is

  1. $60^{0}$
  2. $90^{0}$
  3. $120^{0}$
  4. $135^{0}$
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C Correct answer
Explanation

Given cot(A/2) : cot(B/2) : cot(C/2) = 1 : 4 : 15. This implies tan(A/2) : tan(B/2) : tan(C/2) = 1 : 1/4 : 1/15. Using the identity for triangles, this leads to angles where the largest angle is 120 degrees.

AI explanation

Using the half-angle formula, cot of an angle divided by 2 equals the semi-perimeter s minus the opposite side divided by the inradius r. Given the ratio 1 to 4 to 15, let the values of s minus a, s minus b and s minus c be k, 4k and 15k respectively. Adding these gives 3s minus the sum of the sides, which equals s, so s equals 20k. The side c is s minus 15k, giving 5k, the side b is 16k, and the side a is 19k. Since side a is the largest, angle A is the largest, and using the cosine rule yields cos A equals (19k squared plus 16k squared minus 5k squared) divided by (2 times 19k times 16k), which equals negative one half. Therefore, the greatest angle is 120 degrees.

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