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Math Help - trigo 3D application

  1. #1
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    trigo 3D application

    I have done (a)(b)(c) but fail to do (d). The book's answer of (d) is 67.2 degree.

    Three points A, B and C on horizontal ground are such that AB = 10m, BC = 21m and cos \angle ABC = \frac 3 5 . The point P lies on BC and \angle APB = 90 degree. Calculate

    (a) the length of AC
    (b) the length of AP

    A vertical pole VA of height 19m is placed at A. Calculate, to the nearest tenth of a degree, the acute angle between

    (c) VB and the horizontal
    (d) the plane VBC and the horizontal

    Here's my working of (d) but my answer is wrong.

    let VB = a , VC = b
    a^2 = 19^2 + 10^2 = 461
    b^2 = 19^2 + 17^2 = 650

    consider triangle VBC,
    cos B = \frac {a^2 + 21^2 + b^2}{41a}= \frac{252}{41 \sqrt{461}}
    let VX be the height of triangle VBC and VX perpendicular to BC,
    VX = a\sin B = 20.57
    angle between VBC and the horizontal = \angle VXA = sin ^{-1} \frac {19}{20.57} = 67.45 degree (wrong)

    Thanks in advance!
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  2. #2
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    Hello cakeboby
    Quote Originally Posted by cakeboby View Post
    I have done (a)(b)(c) but fail to do (d). The book's answer of (d) is 67.2 degree.

    Three points A, B and C on horizontal ground are such that AB = 10m, BC = 21m and cos \angle ABC = \frac 3 5 . The point P lies on BC and \angle APB = 90 degree. Calculate

    (a) the length of AC
    (b) the length of AP

    A vertical pole VA of height 19m is placed at A. Calculate, to the nearest tenth of a degree, the acute angle between

    (c) VB and the horizontal
    (d) the plane VBC and the horizontal

    Here's my working of (d) but my answer is wrong.

    let VB = a , VC = b
    a^2 = 19^2 + 10^2 = 461
    b^2 = 19^2 + 17^2 = 650

    consider triangle VBC,
    cos B = \frac {a^2 + 21^2 + b^2}{41a}= \frac{252}{41 \sqrt{461}}
    let VX be the height of triangle VBC and VX perpendicular to BC,
    VX = a\sin B = 20.57
    angle between VBC and the horizontal = \angle VXA = sin ^{-1} \frac {19}{20.57} = 67.45 degree (wrong)

    Thanks in advance!
    I'm not quite sure why you think you need the point X. The height of triangle VBC is VP, since this line (like AP) is perpendicular to BC. (See attached diagram.)

    So the angle you want is \angle VPA = \tan^{-1}\left(\frac{VA}{AP}\right)=\tan^{-1}\left(\frac{19}{8}\right)=67.2^o.

    Grandad
    Attached Thumbnails Attached Thumbnails trigo 3D application-untitled.jpg  
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  3. #3
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    Quote Originally Posted by Grandad View Post
    Hello cakebobyI'm not quite sure why you think you need the point X. The height of triangle VBC is VP, since this line (like AP) is perpendicular to BC. (See attached diagram.)

    So the angle you want is \angle VPA = \tan^{-1}\left(\frac{VA}{AP}\right)=\tan^{-1}\left(\frac{19}{8}\right)=67.2^o.

    Grandad
    Thanks for the reply but \tan ^{-1}\frac {19}{8} = 63.43
    I though about the same problem before posting the question but 63.43 didn't agree with the book's answer so I tried finding another method by creating the point X to find the angle. However I still got the wrong answer.
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  4. #4
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    Hello cakeboby
    Quote Originally Posted by cakeboby View Post
    Thanks for the reply but \tan ^{-1}\frac {19}{8} = 63.43...
    Not on my calculator.

    Grandad

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  5. #5
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    Quote Originally Posted by Grandad View Post
    Hello cakebobyNot on my calculator.

    Grandad

    Oh, sorry! That's my mistake! Thanks!
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