# Did I do something wrong in this problem?

Consider this: "A boat sails $6$ km West, then $5$ km Northwest. Use trigonometry to find the boat's distance and bearing from its starting point."

To solve this, I first drew on paper a shape sort of like this: I used a protractor and ruler to make sure the lines were to scale and with accurate angles/bearing.

What I did from there was take $\tan^{-1}\frac{5}{6}$ to find the angle of $O$, which was approximately $39.8^{\circ}$. Then I did this calculation to find the length of $OB$:$$\sin 39.8=\frac{5}{x}$$ $$x\sin39.8=5$$ $$x=\frac{5}{\sin 39.8}$$ $$x=7.8$$

So there I have my answer. The distance from the starting point is $7.8$ km and the bearing relative to the starting point is $39.8^{\circ}$. However, when I use my ruler and protractor to get an approximate on paper, I get a wildly different result. I made $1$ cm equal $1$ km on my drawing, so the line $OB$ should be $7.8$ centimeters, but it's a lot closer to $10$ centimetres. And $\angle O$ is approximately $21^{\circ}$ when I use my protractor to verify.

So, did I do something wrong? I know paper will always be inaccurate compared to exact calculations, but the difference seems way too big. Did I go wrong somewhere?

• Draw a perpendicular $AD$ from $A$ on $OB$ and $\angle BAO=90^\circ+45^\circ$ and $BD=AD$ Dec 31, 2013 at 19:09

By the law of cosines $$OB^2 = 6^2+5^2 - 2 \cdot 6\cdot 5 \cos(135^\circ)$$ gives $OB \approx 10.2$.

Note that the coordinates of $B$ is not $[-6,5]$ but $[-(6+5/\sqrt{2}), 5/\sqrt{2}]$

Note: Added as an after thought.

In your calculations, you mistakenly used $5$Km North, instead of $5$Km Northwest. I see where your mistake is.

The arctan function can be used to get one of the angles of a right triangle. Your triangle is not a right triangle.

What you can do instead is construct a point C which lies along line AO and which forms right triangle OCB. You know that angle CAB is 45 degrees because they told you that the direction from A is Northwest. Now use what you know about the special 45-45-90 triangle to find the length of line segments OC and CB. From there you can use arctan.

Well, northwest should go to the upper left at $45^{\circ}$, which means that the obtuse angle is $135^{\circ}$.

Then you can use the Law of Cosines to determine the distance $\overline{OB}$, and the Law of Sines to determine the bearing ($\angle AOB$).

Calculating $\tan^{-1}{5\over6}$ for the angle of $O$ would have been appropriate had the second leg been $5$ km due North, not Northwest.

It's easy to see that $39.8$ degrees is too big for the angle at $O$: The angle at $A$ is $135^\circ$, so the angles at $B$ and $O$ sum to $45^\circ$, and the angle at $B$ is larger than the angle at $O$ since $6\gt5$ (longer sides are opposite larger angles). Consequently, the angle at $O$ must be less than $22.5^\circ$.

If you are in fact being asked to solve this problem using vectors (noting the tag), this journey involves adding the vector $\ \overrightarrow{OA} \ = \ \langle 6 \ \cos 180º \ , \ 6 \ \sin 180º \rangle \$ and the vector $\ \overrightarrow{AB} \ = \ \langle 5 \ \cos 135º \ , \ 5 \ \sin 135º \rangle \$ to produce the vector connecting the starting and ending points of the trip,

$$\overrightarrow{AB} \ = \overrightarrow{OA} \ + \ \overrightarrow{AB} \ = \ \langle 6 \ \cos 180º \ + \ 5 \ \cos 135º \ , \ 6 \ \sin 180º \ + \ 5 \ \sin 135º \rangle \ .$$

From these components, you can then calculate the length of the total vector from

$$\sqrt{ (AB_x)^2 + (AB_y)^2}$$

and the direction (measured counter-clockwise from the positive $\ x-$direction from

$$\tan^{-1} \left( \frac{AB_y}{AB_x} \right) ,$$

adding 180º since the direction is in the second quadrant, but the inverse tangent result will be in the fourth quadrant.