Question

If a=3, b=5, c=7 are the sides of a triangle ABC, then \( \cot A + \cot B + \cot C = \)

• A. \( \frac{15\sqrt{3}}{4} \)
• B. \( \frac{7}{\sqrt{3}} \)
• C. \( \frac{83}{15\sqrt{3}} \)
• D. \( \frac{83\sqrt{3}}{15} \)

Hint:

The formula \( \cot A + \cot B + \cot C = \frac{a^2+b^2+c^2}{4\Delta} \) is a very powerful tool for problems involving the sum of cotangents. Memorizing it can save a lot of time compared to finding each angle individually using the Law of Cosines.

Answer 1

The Correct Option is C

A useful formula for the sum of cotangents in a triangle is:
\( \cot A + \cot B + \cot C = \frac{a^2 + b^2 + c^2}{4\Delta} \), where \( \Delta \) is the area of the triangle.
First, we calculate the area \( \Delta \) using Heron's formula.
The semi-perimeter \( s = \frac{a+b+c}{2} = \frac{3+5+7}{2} = \frac{15}{2} \).
Now we find the terms for the formula:
\( s-a = \frac{15}{2} - 3 = \frac{9}{2} \).
\( s-b = \frac{15}{2} - 5 = \frac{5}{2} \).
\( s-c = \frac{15}{2} - 7 = \frac{1}{2} \).
Area \( \Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{15}{2} \cdot \frac{9}{2} \cdot \frac{5}{2} \cdot \frac{1}{2}} \).
\( \Delta = \frac{\sqrt{15 \cdot 9 \cdot 5}}{4} = \frac{\sqrt{3 \cdot 5 \cdot 9 \cdot 5}}{4} = \frac{3 \cdot 5 \sqrt{3}}{4} = \frac{15\sqrt{3}}{4} \).
Next, we calculate the sum of the squares of the sides.
\( a^2 + b^2 + c^2 = 3^2 + 5^2 + 7^2 = 9 + 25 + 49 = 83 \).
Finally, substitute these values into the formula for the sum of cotangents.
\( \cot A + \cot B + \cot C = \frac{83}{4 \cdot (\frac{15\sqrt{3}}{4})} = \frac{83}{15\sqrt{3}} \).