[Math] Solving a Problem with Bearing

trigonometry

A plane flies 1.3 hr at 110mph on a bearing of 38 degrees. It then
turns and flies 1.5 hr at the same speed on a bearing of 128 degrees.
How far is the plane from its starting point?

To solve this, I have drawn a rough sketch. I have also labeled the appropriate sides at 143 miles (distance = rate * time; 1.3hr * 110mph) and 165 miles (1.5hr * 110mph).

But I am having a difficult time concluding which angle within the sketch is the 90 degree angle to allow me to use a trigonometric function or the Pythagorean theorem to solve it.

Edit: It is really important for me to know how to determine the measure of any angle that can be inferred. Not only because I am expecting the professor to attempt to fool us by providing measures where the triangle in question is not a right triangle, but because I thought I had already mastered this skill previously.

Best Answer

Since you turn a total of $90^{\circ}$ the path of the plane forms a right triangle and the two given paths are the legs.

Using pythagorean $a^2+b^2=c^2$, where here $a=1.3\cdot 110=143$ and $b=1.5\cdot 110=165$, so $c=\sqrt{(1.3\cdot 110)^2+(1.5\cdot 110)^2} = 218.344 \text{ mi}$

Related Solutions

[Math] Trigonometric bearing problem

The boat leaves at a bearing of 31 degrees and ends up north of it's start point. Therefore first corner of the triangle is 31 degrees.

At the buoy it has a bearing of 346 which is NW - ish. This angle is equal to 135 degrees. 1st part of angle is 90-31= 59degrees using construction lines and the alternate angles theorem. The second part of the angle is 346-270=76 degrees. (the bearing minus the East-West construction line) so 76+59=135 degrees.

The last angle is 180-135-31=14 degrees.

I now use Sine rule: x/sin135 = 4.3/sin 14. x = 12.57 - which rounds to 13km.

[Math] Trigonometry Airplane question. Finding bearing and distance.

Cosine Law would be easier, but let's do it your way. Consider the small right triangle (call it $\Delta ABC$), which roughly looks like:

A----B
|   /
|  /
| /
|/
C

Note that $\angle ABC =59^\circ$. Thus, $AB=640\cos(59^\circ)$ and $AC=640\sin(59^\circ)$.

Now consider the large right triangle. Its two legs are $AC$ and $AB+480$. Hence, by the Pythagorean Theorem, we have: $$x=\sqrt{(AC)^2 + (AB+480)^2}=\sqrt{(640\cos(59^\circ)+480)^2 + (640\sin(59^\circ))^2} \approx 978 \text{ miles}$$ Finally, for the angle bearing: $$\theta=\tan^{-1}\left(\frac{AC}{AB+480}\right) = \tan^{-1}\left(\frac{640\sin(59^\circ)}{640\cos(59^\circ)+480}\right) \approx 34^\circ \text{ to get the bearing you subtract that value from 90 to get } S56^\circ W$$

Best Answer

Related Solutions

[Math] Trigonometric bearing problem

[Math] Trigonometry Airplane question. Finding bearing and distance.

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