ABC has vertices A(-1,6), B(2,10) and C(7,-2). Find the measure of each angle of the triangle. Round decimal to the nearest tenth

Angle A = 98.1°

Angle B = 59.5°

Angle C = 22.4°

Step-by-step 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°

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step-by-step explanation: