Multiple choice

In a $\Delta$$ABC,\ O$ is a point inside the triangle such that $\angle OBC=\angle OCA=\angle OAB=15^{0}$ then value of $\cot A+\cot B+\cot C$ is

  1. $2-\sqrt{3}$
  2. $\sqrt{2}-1$
  3. $\sqrt{2}+1$
  4. $2+\sqrt{3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

This is a known property for a point O inside a triangle satisfying the given condition. The sum cot A + cot B + cot C for such a point O where the angles are 15 degrees is 2 + sqrt(3).

AI explanation

Because angles OBC, OCA, and OAB are all 15 degrees, the remaining angle parts are A minus 15 degrees, B minus 15 degrees, and C minus 15 degrees. These remaining parts form the interior angles of triangle OAB, OBC, and OCA respectively. Adding these parts gives A plus B plus C minus 45 degrees, which equals 180 degrees. Since A plus B plus C equals 180 degrees for the main triangle, and using the trigonometric identity for cotangents of angles summing to 180 degrees, the sum cot A plus cot B plus cot C equals cot A cot B cot C. Calculating the specific values for this symmetric configuration yields 2 plus the square root of 3.

Keep practicing — related questions