Find the missing side length and angles of △ABC given that m∠A=56∘, b=9, and c=11. In triangle A B C, side A B is 11 units long, side A C is 9 units long, and side B C is labeled a. Angle A measures 56 degrees.

1) Side a = 9.55

2) Angle B = 51.36°

3) Angle C = 72.67°

Step-by-step explanation:

We would be solving the above questions using Law of Cosines

1) To find Side a using Law of cosines

a² = b² + c² - 2bc × Cos A

a = √(b² + c² - 2bc × Cos A)

a = unknown

b = 9

c = 11

Angle A = 56°

a = √(9² + 11² - 2 × 9 × 11 × Cos 56°

a = 9.55405

Approximately a = 9.55

2) To find Angle B

We would use the Law of Cosines as well

b² = a² + c² - 2ac × Cos B

Cos B = a² + c² - b²/2ac

B = arc cos ( a² + c² - b²/2ac)

a = 9.55

c = 11

b = 9

B = arc cos (9.55² + 11² - 9²/2 × 9.55× 11)

B = 51.36°

3) To find Angle C

We would use the Law of Cosines as well

c² = a² + b² - 2ac × Cos C

Cos C = a² + b² - c²/2ab

C= arc cos ( a² + b² - c²/2ab)

a = 9.55

b = 9

c = 11

C = arc cos (9.55² + 9² - 11²/2 × 9.55× 9)

C= 72.67°

i think this could be a possible answer (1 / 84) -1

good luck! : d