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Math Help - Trigonometry

  1. #1
    Junior Member phgao's Avatar
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    Trigonometry

    I don't understand this question, and need some help greatly. Thanks!

    From the top of a cliff 160m high two buoys are observed. Their bearings are 337 and 308. Their respective angles of depression are 3 and 5. Calculate the distance between the bouys.

    What i don't understand is what the bearings mean + if the angle of depression is 3 then shouldn't the bearing be 360 - 3? I'm confused by this.

    Thank you!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by phgao
    I don't understand this question, and need some help greatly. Thanks!

    From the top of a cliff 160m high two buoys are observed. Their bearings are 337 and 308. Their respective angles of depression are 3 and 5. Calculate the distance between the bouys.

    What i don't understand is what the bearings mean + if the angle of depression is 3 then shouldn't the bearing be 360 - 3? I'm confused by this.

    Thank you!
    Assume a flat Earth.

    The depression angles tell you the horizonal range to each buoy, as:

    \tan(\theta_{depression})=\frac{\mbox{height of cliff}}{\mbox{horizontal range to buoy}}

    Their bearings are the angles between North and the direction of the buoy
    measured in the horizontal plane. So the difference in bearings gives
    the angle between the lines joining the foot of the cliff to each of the
    buoys.

    So now you have two sides of a triangle (the two horizontal ranges), and
    the angle between them (the difference in the bearings). If you now solve
    for the third side of this triangle you will have the distance between the
    two buoys.

    RonL
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  3. #3
    Junior Member phgao's Avatar
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    So i have...

    tan3 = 160/horizontal
    Hor = 3052.98

    tan5=160/hor2
    hor2 = 1828.81

    (why) Is a right angled triangle formed?

    If so, is this right:

    tan29 = distance between bouys/3052.98 ?
    Last edited by phgao; February 5th 2006 at 02:30 AM.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by phgao
    So i have...

    tan3 = 160/horizontal
    Hor = 3052.98

    tan5=160/hor2
    hor2 = 1828.81

    Is a right angled triangle formed?
    No the angle between the arms of the triangle is 29 degrees, you
    will have to use the cosine rule to find the third side of the triangle.

    RonL
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  5. #5
    Junior Member phgao's Avatar
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    Oh i see! Thanks!

    So this:

    c^2 = a^2 + b^2 - 2ab cos C

    So c^2 = 3052.98^2 + 1818.81^2 - 2 * 3052.98 * 1818.81 * cos29
    c = 1705.88 or thereabouts as im using 2dp terms.

    Is that right?

    Also, is there anyway it is possible to do it using only sin/cos/tan and not any "rules" such as the cosine rule?

    Ok, I just did it again using full values ie 160tan85 and 160tan87.
    The answer I got was 1702.550 metres.

    One more thing I still can't visualise the triangle and see how angle of depression relates to bearings... it's a bit confusing for me.
    Last edited by phgao; February 5th 2006 at 02:59 AM.
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by phgao
    Oh i see! Thanks!

    So this:

    c^2 = a^2 + b^2 - 2ab cos C

    So c^2 = 3052.98^2 + 1818.81^2 - 2 * 3052.98 * 1818.81 * cos29
    c = 1705.88 or thereabouts as im using 2dp terms.
    Is that right?
    I thought you had 1828.81-which agrees with my calculations

    Also, is there anyway it is possible to do it using only sin/cos/tan and not any "rules" such as the cosine rule?

    Ok, I just did it again using full values ie 160tan85 and 160tan87.
    The answer I got was 1702.550 metres.

    One more thing I still can't visualise the triangle and see how angle of depression relates to bearings... it's a bit confusing for me.
    We try to isolate the depression angles from the bearings, as otherwise
    we end up with a relatively horrible bit of 3D trig. So as in this case the
    depression angles tell us the length of two of the sides of the triangle
    and the bearings give us the angle between the sides.
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  7. #7
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    Quote Originally Posted by phgao
    c^2 = a^2 + b^2 - 2ab cos C

    So c^2 = 3052.98^2 + 1818.81^2 - 2 * 3052.98 * 1818.81 * cos29
    c = 1705.88 or thereabouts as im using 2dp terms.

    Also, is there anyway it is possible to do it using only sin/cos/tan and not any "rules" such as the cosine rule?

    Ok, I just did it again using full values ie 160tan85 and 160tan87.
    The answer I got was 1702.550 metres.
    Of course there are other ways to get the answer without using the Law of Cosines but they are messier. For this kind of triangle, the Law of Cosines way is the better way.

    Another way. "Vector" way.

    Look at the figure from above, like you are a bird----bird's eyeview. or plan view.
    There are two vectors.
    One is 3053 meters long, at bearing 337deg or at (337 -270 =) 67deg clockwise from the negative side of the x-axis. Call this vector OA.
    The other, vector OB, is 1829 meters long, at bearing 308deg or at (308 -270 =) 38deg clockwise from the negative side of the x-axis.

    Bearing here is angle measured clockwise from the positive side of the y-axis or North-South vertical axis.
    [In comparison, regular/ordinary/everyday angles are measured counterclockwise from the positive side of the x-axis.]

    Point A is the tip of vector OA; point B for OB.
    The distance between point A and point B is vector AB.

    The horizontal component, or x-component, of AB is the difference of the x-components of OA and OB.
    (AB)x = 3053cos(67deg) -1829cos(38deg) = -248.4 meters. -----***
    Negative, meaning, point B is farther west than point A.
    "(AB)x" is read "AB, sub x".

    The vertical component, or y-component, of AB is the difference of the y-components of OA and OB.
    (AB)y = 3053sin(67deg) -1829sin(38deg) = 1684.3 meters. -----***

    Then, by Pythagorean Theorem,
    AB = sqrt[(248.4)^2 +(1684.3)^2] = 1702.5 meters -------------answer.

    Quote Originally Posted by phgao
    One more thing I still can't visualise the triangle and see how angle of depression relates to bearings... it's a bit confusing for me.
    You should be able to visualize the triangle now. It is the triangle AOB above.
    Angle AOB = 67 -38 = 29deg.

    Angle of depression, and angle of elevation, are angles of observations. You are standing on one point, you are observing another point that is away from you and this other point is either above (or higher) or below (or lower) the level of your eyes.
    The angle formed by an imaginary horizontal line passing your eye-level and your line of sight to the other point is the angle of observation.
    It is called angle of depression if the other point is depressed or lower than your eye-level. It is called angle of elevation if the other point is elevated or higher than your eye-level.
    Blah, blah, blah.

    ------------------------
    Zeez, man, the Super Bowl XL starts a few hours from now! STEELERS, go, go, go!!!
    I will go to work, do what I need to do over there, then I'd leave my men and I'd go home to watch SB XL here in my room. It's been 26 years late, Man!

    You think I am excited?
    Last edited by ticbol; February 5th 2006 at 10:09 AM. Reason: Can't help it, Zeez, it is Super Bowl a few hours from now!
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  8. #8
    Junior Member phgao's Avatar
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    :cool:

    Once again thanks to both of you!
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