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  1. #1
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    question

    In a right angled triangle ABC Sin A = 5/13

    calculate without finding the angel A the following

    a) Cos A

    b) Tan A
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  2. #2
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    Quote Originally Posted by jimmybob View Post
    In a right angled triangle ABC Sin A = 5/13

    calculate without finding the angel A the following

    a) Cos A

    b) Tan A
    That means the opposite side is 5 and hypotenuse is 13 because sine is opposite over hypotenuse. That means by Pythagorean theorem the adjacent side is 12. Thus, cos A =12/13 because it is adjacent over opposite. And tangent is 5/12 because tangent is opposite over adjacent.
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  3. #3
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    Quote Originally Posted by jimmybob View Post
    In a right angled triangle ABC Sin A = 5/13

    calculate without finding the angel A the following

    a) Cos A

    b) Tan A
    Hello,

    use the formula:

    \sin^2(x)+\cos^2(x)=1. Thus

    \left(\frac{5}{13}  \right)^2+\cos^2(x)=1 \Longrightarrow \cos(A) = \frac{12}{13}

    to b)

    \tan(x)=\frac{\sin(x)}{\cos(x)}. Thus

    \tan(A)=\frac{5}{12}

    EB
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    That means the opposite side is 5 and hypotenuse is 13 ....
    Hello,

    only to be very exact the opposite side is a multiple of 5 and the hypotenuse is the same multiple of 13....

    EB
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