When finding the difference of two vectors, you must subtract the x and y components separately.

\(\displaystyle \vec{A} - \vec{B} = (-2, 4) - (7, -6)\)

\(\displaystyle \vec{A} - \vec{B} = (-2-7, 4 - (-6))\)

\(\displaystyle \vec{A} - \vec{B} = (-9, 10)\)

To find the difference of two vectors we must consider the x and y components separately.

\(\displaystyle \vec{A} - \vec{B} = (5, 8) - (-3, 10)\)

\(\displaystyle \vec{A} - \vec{B} = (5 - (-3), 8 - 10)\)

\(\displaystyle \vec{A} = \vec{B} = (8, -2)\) 

And then we must correctly graph this vector

When finding the magnitude of the vector, you use either the Pythagorean Theorem by forming a right triangle with the vector in question or you can use the distance formula.  This is much more clear considering the distance vector that the magnitude of the vector is in fact the length of the vector.

The correct image depicting an aeronautical bearing of \(\displaystyle 215^\circ\) is 

This image begins at north, and moves \(\displaystyle 215^\circ\) clockwise from it. 

Three of the given incorrect answers depict \(\displaystyle \textup S35^\circ\textup W\), \(\displaystyle 145^\circ\), \(\displaystyle 325^\circ\). The fourth incorrect answer does not represent a standard bearing convention as it is neither an acute angle, nor in the clockwise direction. That incorrect answer looks like:

The bearing of a point B from a point A in a horizontal plane is defined as the acute angle made by the ray drawn from A through B with the north-south line through A. The bearing is read from the north or south line toward the east or west. Bearing is typically only represented in degrees (or degrees and minutes) rather than radians. To find \(\displaystyle \textup S35^\circ\textup W\), start in the south direction, then move \(\displaystyle 35^\circ\) towards the west:

The other incorrect answer choices provided depict \(\displaystyle \textup S 35^\circ \textup E\), \(\displaystyle \textup N 35^\circ \textup E\), and \(\displaystyle \textup N 35^\circ \textup W\).

This question and its answer choices give you a few clues to work with. First, we need to identify the bearing angle being shown. The options in the answer choices are either \(\displaystyle \textup N 30^\circ \textup E\), \(\displaystyle \textup S 30^\circ \textup W\), or \(\displaystyle \textup S 30^\circ \textup E\). Because the angle begins in the south direction and moves \(\displaystyle 30^\circ\) towards the west, the correct bearing is \(\displaystyle \textup S 30^\circ \textup W\). That means only two of the answer choices could be correct. We now need to understand how the \(\displaystyle 180\) miles per hour corresponds to the problem. Notice that there is no answer choice that has the bearing of \(\displaystyle \textup S 30^\circ \textup W\) and velocity of \(\displaystyle 180\) miles per hour. Rather, we need to choose between \(\displaystyle 60\) miles per hour for \(\displaystyle 3\) hours or \(\displaystyle 50\) miles per hour for \(\displaystyle 3\) hours. Because \(\displaystyle 60\) miles per hour for \(\displaystyle 3\) hours corresponds to \(\displaystyle 180\) (whereas the other option corresponds to only \(\displaystyle 150\)), the correct answer is "A motorboat traveling \(\displaystyle \textup S 30^\circ \textup W\) at \(\displaystyle 60\) miles per hour for \(\displaystyle 3\) hours."

First, let's set up a diagram using the given information. This looks like this:

Next, let's convert this info into a triangle so that we can use trigonometry to solve the problem. We need to calculate that the ship going \(\displaystyle 35\) miles per hour for \(\displaystyle 6\) hours will have traveled \(\displaystyle 35\cdot 6=210\) miles. 

Now we can use trigonometry to determine the missing sides, s and e.

\(\displaystyle \cos 30^\circ=\frac{s}{210}\)

\(\displaystyle 210\cos 30^\circ=s\)

\(\displaystyle 210\cdot \frac{\sqrt3}{2}=181.87=s\)

\(\displaystyle \sin 30^\circ=\frac{e}{210}\)

\(\displaystyle 210\sin 30^\circ=e\)

\(\displaystyle 210\cdot \frac{1}{2}=105=e\)

Therefore the ship has travelled 181.87 miles south and 105 miles east. 

First, let's incorporate the given information into a diagram. Start by labelling the plane's bearing of \(\displaystyle 115^\circ\) along with its velocity 330km. Next, draw a line segment to complete the triangle and determine the measures of the angles of the triangle. We can determine the angle \(\displaystyle 65^\circ=180^\circ-115^\circ\), we constructed the diagram such that there is a right angle, and finally we can find the third angle by taking \(\displaystyle 180^\circ-65^\circ-90^\circ=25^\circ\). 

The question is asking us how far south and how far east the plane is from its starting point, so we need to now use trigonometry to determine the lengths of the missing sides of the triangle. We will call these sides s for the southward distance and e for the eastward distance.

\(\displaystyle \cos65^\circ=\frac{s}{330}\)

\(\displaystyle 330\cos65^\circ=s\)

\(\displaystyle s=139.46\) km

\(\displaystyle \sin65^\circ=\frac{e}{330}\)

\(\displaystyle 330\sin65^\circ=e\)

\(\displaystyle e=299.08\) km

Therefore the airplane is 139.46 km south of its starting point and 299.08 km east of its starting point.

To solve this problem, begin with a diagram, and label all known information. We know that we can label two angles as \(\displaystyle 22^\circ\) and one length as 240. Then, by deductive reasoning, we can label another angle as \(\displaystyle 68^\circ\) because \(\displaystyle 90^\circ-22^\circ=68^\circ\).

Next, we need to use trigonometry to find the answers to each question we're being asked. To find the distance AB, set up

\(\displaystyle \cos68^\circ=\frac{240}{AB}\)

\(\displaystyle AB=\frac{240}{\cos68^\circ}\)

\(\displaystyle AB=640.67\)

Next, we can find the distance BC. There are two ways to do this since we know two angles of the triangle, but either way you need to use the tangent function.

\(\displaystyle \tan68^\circ=\frac{BC}{240}\)

\(\displaystyle 240\tan68^\circ=BC\)

\(\displaystyle 594.02=BC\)

Finally, we are asked to find the bearing from Ship B to A. Be careful, because in the initial problem we are given the Bearing from Ship A to Ship B. While the angle will remain the same, the direction is different because we are starting at a different initial point (B instead of A). With B as your starting point, A is north and west. Therefore the bearing from Ship B to Ship A is \(\displaystyle \textup N22^\circ\textup W\).

Therefore the correct distance between Ships A and B is 640.67 miles, the distance between ships B and C is 594.02 miles, and the bearing from Ship B to Ship A is  \(\displaystyle \textup N22^\circ\textup W\).

Begin by diagramming the given information; you'll see that the three ships create a right triangle. To solve the question, we need to find the unknown angles of the triangle, then frame our answers as the proper bearings. 

First let's solve for the topmost angle in the diagram, which we'll call \(\displaystyle \angle S\).

\(\displaystyle \tan S=\frac{345}{200}\)

\(\displaystyle \tan^-^1\left ( \frac{345}{200} \right )=59.9^\circ=\angle S\)

By alternate interior angles, the other diagrammed angle will also be equal to \(\displaystyle 59.9^\circ\).

Now, put your cursor on Island Pioneer, and see if you need to move north or south, and then east or west to get to Sea Terraformer. You need to move north, then west. Therefore the bearing between Island Pioneer and Sea Terraformer is \(\displaystyle \textup N59.9^\circ\textup W\). Now put your cursor on Sea Terraformer and complete the same process to get to Island Pioneer; you'll go south, then east. Therefore the bearing from Sea Terraformer to Island Pioneer is \(\displaystyle \textup S59.9^\circ\textup E\).