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Math Help - Right Triangle Angle Question

  1. #1
    Newbie SilentEchoes's Avatar
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    Right Triangle/Unit Triangle angle Question

    Hi, I am just trying to wrap my head around a quiz question here. I ca figure it out through deduction but I would really like to understand it. The question gives a right triangle ADB with AD as the leg, AB as the hypotenuse and angle ADB as the right angle. It is asking for the measure of angle DAB knowing only that line segment AD is 2/3 as long as line segment AB.

    What rule or theorem applies and why is that? How do go about this? Thanks for any help, appreciated.
    Last edited by SilentEchoes; May 28th 2013 at 02:15 AM.
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    Re: Right Triangle Angle Question

    You can draw any triangle you like as long as the base is 2/3 of the length of the hypotenuse, so maybe a triangle with a base of 2 units and a hypotenuse of 3 units. Then you evaluate the angle using a trigonometric ratio.
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    Re: Right Triangle Angle Question

    Silent....
    Read the question again to understand it...it gives you the cosine of the angle you are looking to find ...
    it is easy....try to do it.
    if you need further help go here ....Trigonometric functions - Wikipedia, the free encyclopedia
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    Re: Right Triangle Angle Question

    Quote Originally Posted by MINOANMAN View Post
    Silent....
    Read the question again to understand it...it gives you the cosine of the angle you are looking to find ...
    it is easy....try to do it.
    if you need further help go here ....Trigonometric functions - Wikipedia, the free encyclopedia
    That's not what the OP was having trouble with, the OP did not understand how to interpret this question when the lengths of the triangle are not given, but are rather given as a ratio. It needs to be pointed out that the trigonometric ratios are so called BECAUSE they are equal as long as the sides are in the same ratio, which means that ANY triangle with sides in the given ratio will work
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    Re: Right Triangle Angle Question

    Thank you both, that helped seeing/understanding the context of the question but now I am having trouble with the ratios.

    The ratios give the sine, cosine, tangent, etc... of angle Theta but how do you get the measure in degrees? It's like I can see the problem/answer but just cannot get there. :S
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    Re: Right Triangle Angle Question

    If you have \displaystyle \begin{align*} \sin{(\theta)} = a \end{align*} then \displaystyle \begin{align*} \theta = \sin^{-1}{(a)} \end{align*}.
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    Re: Right Triangle Angle Question

    Okay, I FINALLY got it figured out. Thank you so much, so important to understand this stuff before moving on and this one was driving me nuts. =P
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