A dog in an open field runs 12.0 m east and then 26.0 m in a direction 55.0 ∘ west of north.?

A: The start point
B: The point at which after the dog running 12.0 m east from point A.
C: The point at which after the dog running 26.0 m in a direction of N 55.0° W.
D: The point at which 11.0 m south of the original starting point.

Join BD and the diagram is shown below :
https://c1.staticflickr.com/1/255/32610396392_3d6046ca44_o.png

In ΔABD :
BD² = (12.0² + 11.0²) m² (Pythagorean theorem)
BD² = 265 m²
BD = √265 m

tan∠DBA = AD/AB = 11/12
∠DBA = 42.5°

In ΔDBC :
∠DBC = (90° - 55.0°) + 42.5° = 77.5°

DC² = BC² + BD² - 2 BC BD cos∠DBC (cosine law)
DC² = (26.0)² + 265 - 2 × 260 × √265 × cos77.5° m²
DC² = 758 m²

cos∠C = (BC² + DC² - BD²) / (2 × BC × DC) (cosine law)
cos∠C = [(26.0)² + 758 - 265] / [2 × (26.0) × (√758)]
∠C = 35.3°

55.0° - 35.3° = 19.7°
Final direct that the dog runs = S 19.7° E

Using x,y coordinates
1st loacation = 0x + 0y
after walking 12m east location = 12x + 0Y
add -21.3x + 14.91y for the 26 m run and the dog's location = -9.3x + 14.91y
11 m south = (0x - 11y)
to get there the dog must go (+9.3x - 25.91Y) = 27.53m L70.26 = 27. 53 m at an agle of 19.74` east of due south