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Math Help - Triangles!

  1. #1
    ADY
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    Lightbulb Triangles!

    If you have a triangle

    AC = 8cm
    AB = 3cm
    BC = no length given but C = 20 (Base)

    Consider that triangle ABC. Now there is a further line AD ( D is on the base line and is a right angled) AD is perpendicular to BC and B lies between D and C.

    How do i find the length of AD?

    Thank you
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  2. #2
    MHF Contributor
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    Hi

    If I have well understood the situation is the following


    In this case \sin \alpha = \frac{AD}{AC} therefore AD = AC \: \sin \alpha
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  3. #3
    ADY
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    Well understood!

    I make the length AD , 2.74cm is that right?
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  4. #4
    ADY
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    Now, how do i find the angle of ABD and ABC?
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  5. #5
    A riddle wrapped in an enigma
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    Quote Originally Posted by ADY View Post
    Well understood!

    I make the length AD , 2.74cm is that right?
    Correct.

    Quote Originally Posted by ADY View Post
    Now, how do i find the angle of ABD and ABC?
    \sin \angle ABD = \frac{2.74}{3}, so find \arcsin \frac{2.74}{3} to find the measure of \angle ABD.

    \angle ABC is supplementary to \angle ABD.
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  6. #6
    ADY
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    so how do inverse sine? sin^-1 \frac{2.74}{3} = 1.2 ?
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  7. #7
    MHF Contributor
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    Quote Originally Posted by ADY View Post
    so how do inverse sine? sin^-1 \frac{2.74}{3} = 1.2 ?
    How did you get 1.2 by the way .

    to find the inverse sine, from your calculator , press shift , sin , then you will see sin^{-1} appearing on the screen , then you key in (2.74/3) to get 65.97.
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  8. #8
    ADY
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    I dont know what i'm doing wrong?:

    sin^-^1 \frac{2.74}{3} = 1.151
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  9. #9
    ADY
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    I dont know what i'm doing wrong?:

    sin^-1 \frac{2.74}{3} = 1.151
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  10. #10
    ADY
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    Quote Originally Posted by mathaddict View Post
    How did you get 1.2 by the way .

    to find the inverse sine, from your calculator , press shift , sin , then you will see sin^{-1} appearing on the screen , then you key in (2.74/3) to get 65.97.
    Got it! - Your working in Degree's and i'm working in Radians!
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  11. #11
    ADY
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    How do i find the lengths of DC DB and therefore find BC
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  12. #12
    A riddle wrapped in an enigma
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    Quote Originally Posted by ADY View Post
    How do i find the lengths of DC DB and therefore find BC
    Refer to Running-gag's drawing.

    (AC)^2=(AD)^2+(AD)^2

    DC=\sqrt{(AC)^2-(AD)^2}

    {\color{red}DC=\sqrt{8^2-(8\sin 20)^2}}


    (AB)^2=(AD)^2+(DB)^2

    DB=\sqrt{(AB)^2-(AD)^2}

    {\color{red}DB=\sqrt{3^2-(8\sin 20)^2}}

    {\color{red}BC=DC-DB}
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  13. #13
    ADY
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    Masters you are truely excellent at this! Bit off topic have you been doing it long?

    Going back to finding the angle of ABD, thats 66 so is ABC = 180 - 66 = 114 ?

    Thank you !
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  14. #14
    A riddle wrapped in an enigma
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    Quote Originally Posted by ADY View Post
    Masters you are truely excellent at this! Bit off topic have you been doing it long?

    Going back to finding the angle of ABD, thats 66 so is ABC = 180 - 66 = 114 ?

    Thank you !

    Yes....rounded off.
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  15. #15
    ADY
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    Do you make the area = 8.2cm^2
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