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Math Help - Advanced Trigonometry Homework

  1. #1
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    Advanced Trigonometry Homework

    Hi,
    I am new to this forum, and really urgently need some help with this homework, as my teacher refused to help me. Please could you set me on the right tracks with the following three questions.

    1. A yacht, Y leaves a harbour, H and travels 45km due North until it reaches a marker buoy B. At B, the yacht turns on a bearing of 290 degrees from B and travels for a further 56km until it reaches a lighthouse L. at L it turns back again and travels back in a straight line to H.

    Calculate
    A) the total distance travelled.
    b. the bearing of L from H
    c. the shortest distance between y and b on the return journey from L to H

    2. A man walks from his home H, on a bearing of 060 degrees for 3.2miles until he reaches his friend's house F. At F, he turns onto a bearing of 280 degrees and travels a further 4.1 miles to his sisters house S. At S he turns again, and walks in a straight line home.

    Calculate

    A) the distance he walks
    B) the bearing of H from S
    the shortes distance between the man and F on the mans return journey from S to H


    Thanks for your help in advance,

    Dan
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by danw2007 View Post
    Hi,
    I am new to this forum, and really urgently need some help with this homework, as my teacher refused to help me. Please could you set me on the right tracks with the following three questions.

    1. A yacht, Y leaves a harbour, H and travels 45km due North until it reaches a marker buoy B. At B, the yacht turns on a bearing of 290 degrees from B and travels for a further 56km until it reaches a lighthouse L. at L it turns back again and travels back in a straight line to H.

    Calculate
    A) the total distance travelled.
    b. the bearing of L from H
    c. the shortest distance between y and b on the return journey from L to H

    Thanks for your help in advance,

    Dan
    See diagram.

    Angle LBH = 110 degrees.

    Use law of cosine to find LH

    (LH)^2=(LB)^2+(BH)^2-2(LH)(BH)\cos 110

    LH \approx 82.97 km

    Add the segment lengths:

    45 + 56 + 82.97 = 183.97 km traveled.

    Now use law of sine to find angle LHB

    \frac{\sin 110}{82.97}=\frac{\sin \angle LHB}{56}

    82.97 \sin \angle LHB=56 \sin110

    [ \sin \angle LHB=\frac{56 \sin 110}{82.97}

    \sin \angle LHB \approx .634238722

    \sin^{-1}(.634238722) \approx 39.36^{\circ}

    Bearing from H to L is 360 - 39.36 = 320.64 degrees.

    Bearing from L to H is 320.64 - 180 = 140.64 degrees.

    I'll let you work on the perpendicular distance (shortest distance) from Y to B. See what you can do.

    The logic is the same for the second problem.
    Attached Thumbnails Attached Thumbnails Advanced Trigonometry Homework-cosine1.bmp  
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