Can someone please do all of these im marking brainliest !!!

Yes ofc, but I don’t see anything like a question or a picture of anything.

9514 1404 393

Step-by-step explanation:

These are all Law of Cosine problems. A generic expression for the length of side 'c' opposite angle C, which is defined by sides 'a' and 'b' is ...

  c² = a² +b² -2ab·cos(C)

The square root of this gives the side length:

  c = √(a² +b² -2ab·cos(C))

Rearranging the equation, we can obtain an expression for the angle C.

  C = arccos((a² +b² -c²)/(2ab))

These two formulas are used to solve the offered problems.

__

1) AC = √(13² +14² -2·13·14·cos(109°)) ≈ √483.506

  AC ≈ 22.0

__

2) BC = √(7² +10² -2·7·10·cos(123°)) ≈ √225.249

  BC ≈ 15.0

__

3) ∠B = arccos((24² +28² -14²)/(2·24·28)) = arccos(1164/1344)

  ∠B ≈ 30.0°

__

4) ∠B = arccos((6² +9² -14²)/(2·6·9)) = arccos(-79/108)

  ∠B ≈ 137.0°