JEE Main Maths: Trigonometry - Properties of Triangles questions with solutions

Q1. In triangle ABC, if b² + c² = 1999 * a², find the value of (cot B + cot C) / cot A.

1. 1/999
2. 1/1999
3. 999
4. 1999

Answer: 1/999

Using the formula cot X = (sum of squares of adjacent sides - opposite side squared) / (4 * area): cot B + cot C = [(a²+c²-b²) + (a²+b²-c²)] / (4*Delta) = 2a² / (4*Delta). And cot A = (b²+c²-a²)/(4*Delta). So (cot B + cot C)/cot A = 2a² / (b²+c²-a²). Substituting b²+c² = 1999*a²: denominator = 1999*a² - a² = 1998*a². Ratio = 2a² / (1998*a²) = 1/999.

Q2. In triangle ABC, medians AD and CE have lengths 18 and 27 respectively, and AB = 24. The line CE is extended to meet the circumcircle of triangle ABC at point F. G is the centroid of the triangle. Which of the following statements are correct?

1. b = 3 * sqrt(70)
2. a = 6 * sqrt(31)
3. Area of triangle AGE = (27/4) * sqrt(55)
4. EF = 16/3

Answer: b = 3 * sqrt(70)

Median from A (to midpoint D of BC): mₐ² = (2b² + 2c² - a²)/4 = 324, so 2b² + 2*576 - a² = 1296 => 2b² - a² = 144...(1). Median from C (to midpoint E of AB): m_c² = (2a² + 2b² - c²)/4 = 729, so 2a² + 2b² - 576 = 2916 => 2a² + 2b² = 3492 => a² + b² = 1746...(2). From (1): 2b² - a² = 144 => a² = 2b² - 144. Substituting into (2): 2b² - 144 + b² = 1746 => 3b² = 1890 => b² = 630 => b = sqrt(630) = 3*sqrt(70). Then a² = 1746 - 630 = 1116, a = sqrt(1116) = 6*sqrt(31). Both A and B are correct. For option D: G divides CE in 2:1, so CG = 18, GE = 9. By power of a point or intersecting chords: GE * GF = GA * GD. GA = (2/3)*18 = 12, GD = 6. GE*GF = 12*6 = 72. GF = 72/9 = 8. EF = GE + GF = 9 + 8 = 17, not 16/3. So D is wrong.