Answer:
Angle A = 98.1°
Angle B = 59.5°
Angle C = 22.4°
Explanation:
When given vertices (x1,y1) ,(x2,y2), we would be using the formula:
√(x2 - x1)² + (y2 - y1)²
ABC has vertices A(-1,6), B(2,10) and C(7,-2)
Side AB = A(-1,6), B(2,10)
√(x2 - x1)² + (y2 - y1)²
= √(2-(-1))² + (10 - 6)²
= √(3² + 4²)
= √9+16
= √25
= 5
Side BC = B(2,10), C(7,-2)
√(x2 - x1)² + (y2 - y1)²
√(7-2)² + (-2 - 10)²
= √5² +(-12)²
= √(25 + 144)
= √169
= 13
Side AC = A(-1,6), C(7,-2)
√(x2 - x1)² + (y2 - y1)²
√(7-(-1))² + (-2 - 6)²
= √6² +(-8)²
= √(64 + 64)
= √128
= 11.314
To find the measure of each angle we would be using the cosine rule
Where
a² = b² + c² - 2bc × Cos A
b² = a² + c² - 2ac × Cos B
c² = a² + b² - 2ab × Cos C
Sketching a triangle we find out that
Side AB = c = 5
Side AC = b = 11.314
Side BC = a = 13
Hence
1) for Side BC = a
a² = b² + c² - 2bc × Cos A
Cos A = b² + c² - a²/ 2bc
Cos A = 11.314² + 5² -13²/(2 × 11.313 × 5)
A = arc cos[(11.314² + 5² -13²)/(2 × 11.313 × 5)]
A = 98.13°
To the nearest tenth = 98.1°
2) for Side AC = b
b² = a² + c² - 2ac × Cos B
Cos B = a² + c² - b²/ 2ac
Cos B = 13² + 5² - 11.314/2 × 13 × 5
B = arc cos[ (13² + 5² - 11.314)/(2 × 13 × 5)]
B = 59.49°
To the nearest tenth = 59.5°
3) for Side AB = c
c² = a² + b² - 2ab × Cos C
Cos C = a² + b² - c²/ 2ab
Cos C = 13² + 11.314² - 5²/ 2 × 13 × 11.314
C = arc cos[(13² + 11.314² - 5²)/ (2 × 13 × 11.314)]
C = 22.38°
To the nearest tenth = 22.4°