Question

The vertice of triangle ABC are A(8,4), B(6,3) and C(5-5). Find the angle between the side AB and BC.​

Answer 1

Answer:

Step-by-step explanation:

Use law of cosines

c² = a² + b² - 2abcosθ

cosθ = (c² - a² - b²) / -2ab

c = length of AC = √(8 - 5)² + (4 - (-5))² = √90

b = length of AB = √(8 - 6)² + (4 - 3)² = √5

a = length of BC = √(6 - 5)² + (3 - (-5))² = √65

cosθ = (√90² - √65² - √5²) / -2√65√5

cosθ = 20 / -36.05551...

cosθ = - 0.5547001

θ = 123.69°