Trigonometry word problem, I have an exam tomorrow PLEASE HELPP

Jun 2018
1
0
Canada
[FONT=&quot] At noon two cars travel away from the intersection of two country roads that meet at a 34 degree angle. Car A travels one of the roads at 80km/h and car B travels along the other road at 100km/h. Two hours later, both cars spot a jet in the air between them. The angle of depression from the jet to car a is 20 degrees and the distance between the jet and the car is 100km. Determine the distance between the jet and Car B. [/FONT]
 
Apr 2018
209
140
Tempe, AZ
[FONT="] At noon two cars travel away from the intersection of two country roads that meet at a 34 degree angle. Car A travels one of the roads at 80km/h and car B travels along the other road at 100km/h. Two hours later, both cars spot a jet in the air between them. The angle of depression from the jet to car a is 20 degrees and the distance between the jet and the car is 100km. Determine the distance between the jet and Car B. [/FONT]
There is not enough information given to solve this. The jet is somewhere on a circle above car A. Without more information you can't say how far it is from car B.
 
Dec 2011
2,313
914
St. Augustine, FL.
I would use coordinate geometry where the units are miles. The road along which car A travels will lie along the \(\displaystyle x\)-axis with the intersection of the two roads at the origin. Car A will travel in the positive direction. The road long which car B travels will lie along the line:

\(\displaystyle y=\tan\left(34^\circ\right)x\)

And Can B travels into quadrant I. After 2 hours of travel at the given constant speeds, the coordinates of the cars are as follows:

Car A: \(\displaystyle (160,0)\)

Car B: \(\displaystyle (200\cos\left(34^\circ\right), 200\sin\left(34^\circ\right))\)

And so the distance \(\displaystyle d\) between the cars is:

\(\displaystyle d=\sqrt{(200\sin\left(34^\circ\right)-0)^2+((200\cos\left(34^\circ\right)-160)^2}=40\sqrt{41-40\cos\left(\frac{17\pi}{90}\right)}\approx111.989264469757688\)

Now, assuming the jet is in the vertical plane containing the two cars, the distance \(\displaystyle A\) along the ground from car A to the point on the ground directly below the jet

\(\displaystyle A=100\cos\left(20^{\circ}\right)\)

And so the altitude \(\displaystyle h\) of the jet is:

\(\displaystyle h=100\sin\left(20^{\circ}\right)\)

Now, we have:

\(\displaystyle A+B=d\implies B=d-A\)

Then the distance from car B to the jet is then:

\(\displaystyle \sqrt{h^2+B^2}\approx\sqrt{18.020002391166855^2+34.20201433256687^2} \approx38.65874119242333\quad\checkmark\)
 
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Apr 2018
209
140
Tempe, AZ
Now, assuming the jet is in the vertical plane containing the two cars,
I suppose that is one way to interpret "between", but it surely isn't a given...