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Math Help - Trigonometry Perpendicular line help!?

  1. #1
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    Trigonometry Perpendicular line help!?

    8. In a triange ABC, AB = 4.8cm, BC = 3.8 cm and A ^B C =42 degrees

    P is the foot of the perpendicular from A to BC.
    i. Calculate the area of triangle ABC


    ii. Deduce the length of AP.


    _______________________________________


    Okay, I don't understand what they mean by "P is the foot of the perpendicular from A to BC"
    And how would i go about deducing the answer? Help
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  2. #2
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    Re: Trigonometry Perpendicular line help!?

    did you make a sketch ... ?

    from vertex A to side BC , AP is the altitude from A to BC.

    also ...

    ... and A ^B C =42 degrees
    what angle is this?
    Attached Thumbnails Attached Thumbnails Trigonometry Perpendicular line help!?-trianglealtitude.png  
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  3. #3
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    Re: Trigonometry Perpendicular line help!?

    Okay thanks.

    soo I worked out the area to be 61.02 usuing 1/2*a*b*sin C

    now for part II i'm not sure what to do.. I know the angle is 90. can u give me a hint or guide me a bit? :S
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    MHF Contributor MarkFL's Avatar
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    Re: Trigonometry Perpendicular line help!?

    Your area is 10 times greater than it should be. To find the altitude, simply use:

    \sin(42^{\circ})=\frac{\bar{AP}}{\bar{AB}}
    Last edited by MarkFL; October 22nd 2012 at 07:14 PM.
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  5. #5
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    Re: Trigonometry Perpendicular line help!?

    But i don't know what AP is!
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  6. #6
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    Re: Trigonometry Perpendicular line help!?

    oKAY I used Area= 1/2*base*height

    and then i rearranged theformula to get the height as 3.7

    correct?
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  7. #7
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    Re: Trigonometry Perpendicular line help!?

    You have:

    A=\frac{1}{2}\bar{BC}\cdot\bar{AB}\sin(42^{\circ})

    and:

    A=\frac{1}{2}\bar{BC}\cdot\bar{AP}

    Hence:

    \bar{AP}=\bar{AB}\sin(42^{\circ}) (which you would get from my previous post)

    \bar{AP}\approx3.2\text{ cm}
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