The bearing angle exists 53.49°.
A bearing of 36° corresponds to the corresponding angle of
θ = 90 -36 = 54°
The (x, y) values for the position of the ship after achieving its first heading exist:
[tex]x_1[/tex] = 12.1 Cos 54° = 7.11 m
[tex]y_1[/tex] = 12.1 Sin 54° = 9.78 m
Our second displacement exists a simple 4.6 mi due east, that exists, the positive x-direction. The components exist therefore
[tex]x_2[/tex] = 6.1 mi
[tex]y_2[/tex] = 0 mi
To find the total displacement from the port, we'll add these two vectors' components and use the distance formula:
Δx [tex]= x_1+x_2[/tex]
= 7.11 mi + 6.1 mi = 13.21 mi
Δy [tex]=y_1+y_2[/tex]
= 9.78 mi + 0 mi = 9.78 mi
[tex]$r=\sqrt{(x_{total})^2+(y_{total})^2}[/tex]
[tex]=\sqrt{(13.21mi)^2+(9.78mi)^2}[/tex]
= 16.43 mi
The direction of the displacement vector exists given by
tan θ = Δy/Δx
θ = arc tan (Δy/Δx)
= arc tan (9.78/13.21)
= 36.51°
The question requested for the bearing angle, which exists just this angle subtracted from 90°:
90° - 36.51° = 53.49°.
Therefore, the bearing angle exists 53.49°.
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