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Math Help - In a triangle-sine of two angles are given .....find cos of 3rd angle

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    In a triangle-sine of two angles are given .....find cos of 3rd angle

    the sines of two angles of triangle are 5/13 and 99/101 ...find cosine of third angle?
    1)245/313
    2)255/313
    3)735/1313
    4)765/1313
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  2. #2
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    Re: In a triangle-sine of two angles are given .....find cos of 3rd angle

    Hello, gurparwaan!

    \text{The sines of two angles of triangle are }\tfrac{5}{13}\text{ and }\tfrac{99}{101}
    \text{Find the cosine of third angle.}

    . . (1)\;\frac{245}{1313} \quad (2)\;\frac{255}{1313} \quad (3)\;\frac{735}{1313} \quad (4)\;\frac{765}{1313}

    \text{Let: }\sin A = \frac{5}{13}\;\hdots\text{ then: }\cos A = \frac{12}{13}

    \text{Let: }\sin B = \frac{99}{101}\;\hdots\text{ then: }\cos B = \frac{20}{101}}


    \text{We have: }\:A + B + C \:=\:180^o \quad\Rightarrow\quad C \:=\:180^o - (A+B)

    \text{Then: }\:\cos C \;=\;\cos[180^o - (A+B)] \;=\;-\cos(A+B)

    \text{Hence: }\:\cos C \;=\;-\cos A\cos B + \sin A\sin B

    . . . = . . . . . . =\;-\left(\frac{12}{13}\right) \left(\frac{20}{101}\right) + \left(\frac{5}{12}\right) \left(\frac{99}{101}\right)

    . . . . . . . . . . =\;-\frac{240}{1313} + \frac{495}{1313}

    . . . . . . . . . . =\;\;\;\frac{255}{1313} \;\;\hdots\;\text{ answer (2)}

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