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Math Help - Complex (for me) Right Triangle

  1. #1
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    Complex (for me) Right Triangle

    Hello; essentially, Iím new to trig, and cannot figure out the problem below. I am able to manage what I would consider straight forward Right Triangles, but this one has a wee bit on the end, that I do not know how to sort out.
    I believe using SOH-CAH-TOA (new term for me) is the right method to solve this, as (again I believe) Sine requires 2-angle/sides, and Cosine requires 1-angle/sideÖ?
    Best I was able to manage was figure out Angle D (180-90-30 = 60).
    Additionally, I thought that the red forward arrows (hash-marks) meant that two-sides were equal, but Iím not sure in this case. I would appreciate knowing what these marks indicate.
    I would appreciate any help on how to approach and solve this problem. Iím one of those people who require a good example to draw upon for future reference.
    Thanks in advance for the forums help!



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  2. #2
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    Quote Originally Posted by Lightheaded View Post
    Hello; essentially, Iím new to trig, and cannot figure out the problem below. I am able to manage what I would consider straight forward Right Triangles, but this one has a wee bit on the end, that I do not know how to sort out.
    I believe using SOH-CAH-TOA (new term for me) is the right method to solve this, as (again I believe) Sine requires 2-angle/sides, and Cosine requires 1-angle/sideÖ?
    Best I was able to manage was figure out Angle D (180-90-30 = 60).
    Additionally, I thought that the red forward arrows (hash-marks) meant that two-sides were equal, but Iím not sure in this case. I would appreciate knowing what these marks indicate.
    I would appreciate any help on how to approach and solve this problem. Iím one of those people who require a good example to draw upon for future reference.
    Thanks in advance for the forums help!



    You can use similar triangles to solve this.

    tan(30) = opp/adj = 7/(BC). tan(30) is a common value on the unit circle and it's sqrt(3)/3

    BC = \frac{7}{tan(30)} = \frac{7\sqrt(3)}{3}

    \frac{7\sqrt(3)}{3} = 7/x and so x = \sqrt3

    now you can work out the two hypotenuses with Pythagoras
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  3. #3
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    Thank you for your reply!


    Keep in mind, Iím still relatively new at this, but letís see what happens...


    Step 1: Find value for BC


    7/tan(30) = 7 sqrt(3) = 12.1243


    BC = 12

    ---------------------------------

    Step 2: Find value for BD

    BD = 12^2 + 7^2 = sqrt(193) = 13.8924

    BD = 14

    ---------------------------------

    Now I gather that the red arrows indicate that each side of the smaller right triangle are equal at 1mm each. That being the case, then Iím confused about the red arrows on the larger right angle...as that would then suggest that BC is also equal to BD?

    Regardless, Iíll move forward, and hopefully someone can help me understand further.

    We now end with:

    Step 3: AB + BD = AD


    1 + 14 = 15

    AD = 15 mm

    Is this correct?
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