#### laxshiny24

[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]

#### Walagaster

[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.

#### IamSujith

I found an answer of 38.7Km to 3s.f any one else.

1 person

#### MarkFL

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:

$\displaystyle \sqrt{h^2+B^2}\approx\sqrt{18.020002391166855^2+34.20201433256687^2} \approx38.65874119242333\quad\checkmark$

1 person

#### Walagaster

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...