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Math Help - elevation

  1. #1
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    elevation

    From a point A south of the vertical tower , the angle of elevation of the top of the tower is observed to be 34 degree . From A , an observer walked a distance of 270 m on a bearing 070 degree and found that his bearing from the tower is 135 degree . Cal culate the height of the tower to the nearest metre .
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  2. #2
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    Quote Originally Posted by thereddevils View Post
    From a point A south of the vertical tower , the angle of elevation of the top of the tower is observed to be 34 degree . From A , an observer walked a distance of 270 m on a bearing 070 degree and found that his bearing from the tower is 135 degree . Cal culate the height of the tower to the nearest metre .
    make a sketch ...

    let point T be the tower's position and point B be the position the observer walks to from point A.

    triangle TAB is formed ...

    angle A = 70

    angle B = 65

    angle T = 45

    AB = 270 m

    use the law of sines to find AT, then use the tangent ratio to find the tower's height ...

    tan(34) = h/(AT)
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  3. #3
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    This is a 3D problem. I did one similar to it in this thread: http://www.mathhelpforum.com/math-he...g-problem.html . But there are some differences.

    First you'll need to imagine a bird's eye view of the ground. See attachment #1. You've got a triangle ABT. Remember that bearings are measured with 0 = N. Angle T is 45, angle A is 70, and angle B is 65. Use the Law of sines to find side AT:
    \frac{\sin 45^{\circ}}{270} = \frac{\sin 65^{\circ}}{AT}

    Why AT? Because you need that length to find the height of the tower at T. See attachment #2. (Actually, the placement of T here is wrong -- it should be at the base of the tower, not the top. Sorry about that.) Use the tangent ratio to find AT.
    \tan 34^{\circ} = \frac{h}{AT}


    01

    EDIT: Beaten to it by skeeter...
    Attached Thumbnails Attached Thumbnails elevation-tower_3d.jpg   elevation-tower_3d2.jpg  
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  4. #4
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    thanks .. but i don seem to be able to see from my diagram to get the 3 angles . Maybe my sketch is wrong and i am pretty confused with the diagram . Do u mind posting up ur sketch? Its ok if it's too much of work . Thank you once again .
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  5. #5
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    here is the basic bird's eye view ... you should be able to find the value of every angle from this diagram.
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  6. #6
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    Quote Originally Posted by yeongil View Post
    This is a 3D problem. I did one similar to it in this thread: http://www.mathhelpforum.com/math-he...g-problem.html . But there are some differences.

    First you'll need to imagine a bird's eye view of the ground. See attachment #1. You've got a triangle ABT. Remember that bearings are measured with 0 = N. Angle T is 45, angle A is 70, and angle B is 65. Use the Law of sines to find side AT:
    \frac{\sin 45^{\circ}}{270} = \frac{\sin 65^{\circ}}{AT}

    Why AT? Because you need that length to find the height of the tower at T. See attachment #2. (Actually, the placement of T here is wrong -- it should be at the base of the tower, not the top. Sorry about that.) Use the tangent ratio to find AT.
    \tan 34^{\circ} = \frac{h}{AT}


    01

    EDIT: Beaten to it by skeeter...

    I can follow up still \tan 34^{\circ} = \frac{h}{AT}

    why not \sin 34^{\circ} = \frac{h}{AT}

    How come the placement of T is at the bottom ?
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  7. #7
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    Quote Originally Posted by thereddevils View Post
    I can follow up still \tan 34^{\circ} = \frac{h}{AT}

    why not \sin 34^{\circ} = \frac{h}{AT}

    How come the placement of T is at the bottom ?
    Use the definition of Sine:

    \sin(angle)=\dfrac{leg}{hypotenuse}

    Since AT is the hypotenuse this distance must be the denominator. (By the way: The confusion is caused by yeongil's labelling of the first sketch: In a bird eye's view the points T and F are of course on the same place, but ...)

    Maybe it's easier to label the footpoint of the tower by F. Then you'll get:

    \tan 34^{\circ} = \frac{h}{AF}
    Attached Thumbnails Attached Thumbnails elevation-turm_mitsinussatz.png  
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  8. #8
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    Yes, apologies again for the confusion. I need to stress that T is supposed to be at the base of the tower, not the top. The reason T is not at the top is because when I found AT from the 1st diagram, AT was the distance from A to the tower along the ground, so that corresponds to T being the base of the tower in the 2nd diagram.

    Therefore, we cannot use \sin 34^{\circ} = \frac{h}{AT}. We don't even know the distance from point A to the top of the tower (hypotenuse) - just the distance along the ground (adjacent).


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