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Thread: Trigonometry word problem, I have an exam tomorrow PLEASE HELPP

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    Trigonometry word problem, I have an exam tomorrow PLEASE HELPP

    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.
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    Re: Trigonometry word problem, I have an exam tomorrow PLEASE HELPP

    Quote Originally Posted by laxshiny24 View Post
    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.
    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.
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    Re: Trigonometry word problem, I have an exam tomorrow PLEASE HELPP

    I found an answer of 38.7Km to 3s.f any one else.
    Thanks from MarkFL
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    Re: Trigonometry word problem, I have an exam tomorrow PLEASE HELPP

    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:

    $\displaystyle d=\sqrt{(200\sin\left(34^\circ\right)-0)^2+((200\cos\left(34^\circ\right)-160)^2}=40\sqrt{41-40\cos\left(\frac{17\pi}{90}\right)}\approx111.989 264469757688$

    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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    Re: Trigonometry word problem, I have an exam tomorrow PLEASE HELPP

    Quote Originally Posted by MarkFL View Post

    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...
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