how to find the magnitude and direction of the displacement vector OB when X=7,Y=6?

Question

In a desert exercise a tank travels x km on a bearing of 050degree from a base O then y km on a bearing of 140degree to a bunker B.

please show me how to draw the diagram as well.

Answer 1

Bearings start with 0 at north and go clockwise around the circle.

First find the magnitude.

When the tank changed bearing from 050° to 140° it made a turn of (140 - 50) = 90°. So we have a right triangle. OB is the hypotenuse and x and y are the legs.

Let
d = distance OB

d = √(x² + y²) = √(7² + 6²) = √(49 + 36) = √85

h = horizontal distance
v = vertical distance

h = 7sin50° + 6sin140° ≈ 9.2190368
v = 7cos50° + 6cos140° ≈ -0.0967533

Bearing = 90° - arctan(v/h) ≈ 90° - arctan(-0.0104949)

Bearing = 90° + 0.6012946° = 90.6012946°

Answer 2

The answer is the sum of the two travels. For each travel, you have been given magnitude and direction. Convert to Cartesian coordinates. Sum the travels in that form, then convert back to polar coordinates for your answer.

direction = arctan((7 km sin(50 deg) + 6 km sin(140 deg)) / (7 km cos(50 deg) + 6 km cos(140 deg))).

magnitude = sqrt(((7 km sin(50 deg) + 6 km sin(140 deg)) squared + (7 km cos(50 deg) + 6 km cos(140 deg)) squared)

Answer 3

magnitude = sqrt(49+36) =sqrt(85)
Direction tan @= 6/7 so @=0.7086 rad

Answer 4

Magnitude: 9
Direction: 090 degree
Please give me best answer thanks!