How do you solve this physics problem

Question

During a rodeo, a clown runs 8.0 m north, turns 55° north of east, and runs 3.5 m. Then, after waiting for the bull to come near, the clown turns due east and runs 5.0 m to exit the arena. What is the clown’s total displacement?

Answer 1

Translate "8 Meters Directed North from O" (the original position) to the Vector 0i + 8j.

Translate "3.5 Meters Directed 55° North of East from the Endpoint of the Vector 0i + 8j" to the

Vector (3.5×cos 55°)i + (3.5×sin 55°)j.

Translate "5 Meters Directed Due East from the Endpoint of the Vector (3.5×cos 55°)i + (3.5×sin 55°)j"

to the Vector 5i + 0j.

Stack the Vectors as shown below:

----------------0i---+------------------- 8j

(3.5×cos 55°)i------+ (3.5×sin 55°)j

----------------5i+-----------------------0j.

Now obtain (0+ 3.5cos 55°+5)i + (8+ 3.5sin 55°+0)j

which goes to 7.007517527i + 10.86703216j.

7.007517527i + 10.86703216j calculates to √(7.007517527² + 10.86703216²) Meters

at [Arctangent of (10.86703216/7.007517527)] Degrees.

The total displacement of the clown is then

12.93049457 Meters at 57.18438201°

North of Due East

from his original position.