# Thread: I am having trouble graphing this.

1. ## I am having trouble graphing this.

a plane leaves an airport and travels 624km due east. It then turns toward the north and travels another 326 km. It then turns again less than 180 degrees and travels another 846 km directly back to the airport. Through what angles did it turn?

I get the east at 624 then north from there at 326. But what does it mean turns again less than 180 degrees and travels another 846?

Stuck on this, help if you can, thanks

What does it mean turns again less than 180 degrees and travels another 846?
This means it turns some angle that it tells us leads back to the airport. Since we are northeast of the airport, any angle 0-180 degrees to the east would be moving away from the airport in an eastern direction. So the angle is going to be less than 180 degrees to the left because the plane takes this path back to the airport which is southwest of 624 km east, 326 km north.

The picture for this is a triangle. A base leg 624 km (east), another leg or the height 326 km (north) and a hypotenuse 846 km (west-southwest). To solve you can use your inverse sine

We're looking for angle X at the top of the triangle so

$\displaystyle (\sin{X})= \frac{624}{846}$

$\displaystyle (\cos{X})= \frac{326}{846}$

$\displaystyle (\tan{X})= \frac{624}{326}$

3. only problem is $\displaystyle 624^2 + 326^2 \ne 846^2$ ... something's not right.

. It then turns toward the north and travels another 326 km.
Aha this is where the problem lies. The triangle is not right, the plane turns 'toward the north' not 'due north', so basically your triangle is slanted and we have to solve for the three angles.

5. The law of cosines states the cosine angle A between the sides 846 and 624 is equal to $\displaystyle \frac{b^2+c^2-a^2}{2bc}$ where a is the side opposite angle A, b the side opposite angle B and c the side opposite angle C (I labeled the angles counter-clockwise after A)

Starting by finding the measure of angle A:

$\displaystyle \cos^{-1}(\cos{A})=\cos^{-1}(\frac{b^2+c^2-a^2}{2bc})$

$\displaystyle \cos^{-1}(\cos{A})=\cos^{-1}(\frac{846^2+624^2-326^2}{2(846)(624)})$ so m<A is about 18.9 degrees.

$\displaystyle \cos^{-1}(\cos{C})=\cos^{-1}(\frac{a^2+b^2-c^2}{2ab})$

$\displaystyle \cos^{-1}(\cos{C})=\cos^{-1}(\frac{326^2+846^2-624^2}{2(326)(846)})$ so m<C is about 38.3 degrees.

180-(38.3+18.9)=about 122.8 degrees for angle B

m<A= approx. 18.9 degrees
m<B= approx. 122.8 degrees
m<C= approx. 38.3 degrees

This makes sense in the problems context since it turns north at angle B the angle should be greater than 90 degrees =)

6. Thank you all for the reply. I understand this now.