Class 9 Higher Math Solution Bd [2025]

Solution: AB = √[(4-1)² + (6-2)²] = √(9+16) = √25 = 5 BC = √[(7-4)² + (2-6)²] = √(9+16) = 5 CA = √[(1-7)² + (2-2)²] = √(36+0) = 6

Solution: a=2, b=-5, c=2 Δ = (-5)² - 4×2×2 = 25 - 16 = 9 x = [5 ± √9] / (4) = [5 ± 3]/4 Class 9 Higher Math Solution Bd

(12, ∞) Chapter 2: Algebraic Expressions 2.1 Key Formulas (Memorize!) | Identity | Expansion | |----------|-----------| | (a+b)² | a² + 2ab + b² | | (a-b)² | a² - 2ab + b² | | a² - b² | (a+b)(a-b) | | (a+b)³ | a³ + 3a²b + 3ab² + b³ | | (a-b)³ | a³ - 3a²b + 3ab² - b³ | | a³ + b³ | (a+b)(a² - ab + b²) | | a³ - b³ | (a-b)(a² + ab + b²) | 2.2 Worked Example Q: Factorize: x⁴ + x² + 1 Solution: AB = √[(4-1)² + (6-2)²] = √(9+16)

Since AB = BC = 5, triangle ABC is isosceles. 4.1 Ratios for 0°, 30°, 45°, 60°, 90° | θ | sinθ | cosθ | tanθ | |---|------|------|------| | 0° | 0 | 1 | 0 | | 30° | 1/2 | √3/2 | 1/√3 | | 45° | √2/2 | √2/2 | 1 | | 60° | √3/2 | 1/2 | √3 | | 90° | 1 | 0 | ∞ | 4.2 Example Problem Q: If tanθ = 3/4, find sinθ and cosθ. Class 9 Higher Math Solution Bd

Solution: Let opposite=3k, adjacent=4k Hypotenuse = √[(3k)² + (4k)²] = √(9k²+16k²) = 5k