is a clockwise angle measured from due North. This is a problem, because all of the trigonometric functions are referenced to a counterclockwise angle measured from East.
A bearing of <span>34∘</span> corresponds to a trigonometric angle of <span><span>θ1</span>=<span>90∘</span>−<span>34∘</span>=<span>56∘</span></span>
The (x,y) values for the position of the ship after completing its first heading are:
<span>x=<span>(10.4mi)</span><span>cos<span>(<span>56∘</span>)</span></span></span>
<span>y=<span>(10.4mi)</span><span>sin<span>(<span>56∘</span>)</span></span></span>
The trigonometric angle for the second heading is <span><span>θ2</span>=<span>90∘</span>−<span>90∘</span>=<span>0∘</span></span>
The (x,y) values for the position of the ship after completing its second heading is:
<span>x=<span>(10.4mi)</span><span>cos<span>(<span>56∘</span>)</span></span>+<span>(4.6mi)</span><span>cos<span>(<span>0∘</span>)</span></span>≈10.4mi</span>
<span>y=<span>(10.4mi)</span><span>sin<span>(<span>56∘</span>)</span></span>+<span>(4.6mi)</span><span>sin<span>(<span>0∘</span>)</span></span>≈8.6mi</span>
The distance from port is:
<span>d=<span>√<span><span><span>(10.4)</span>2</span>+<span><span>(8.6)</span>2</span></span></span>≈13.5mi</span>
Its trigonometric angle is:
<span>θ=<span><span>tan<span>−1</span></span><span>(<span>yx</span>)</span></span></span>
<span>θ=<span><span>tan<span>−1</span></span><span>(<span>8.610.4</span>)</span></span></span>
<span>θ≈<span>39.6∘</span></span>
The bearing angle is:
<span><span>θb</span>=<span>90∘</span>−<span>39.6∘</span>=<span>50.4<span>∘</span></span></span>
