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  1. #1
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    trig

    In a right triangle given a=32.8ft and A= 28 deg 30’15” , solve for b and c
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  2. #2
    Senior Member DivideBy0's Avatar
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    Quote Originally Posted by harry View Post
    In a right triangle given a=32.8ft and A= 28 deg 30’15” , solve for b and c
    \left(28+\frac{30}{60}+\frac{15}{360} \right)^\circ =\frac{685 }{24}^\circ

    Using the Sine Rule:

    \frac{a}{\sin A}=\frac{b}{\sin B} = \frac{c}{\sin C}

    \frac{32.8}{\sin(\frac{685}{24})}=\frac{c}{\sin90}

    c \approx 68.65

    Use pythag to get b:

    b=\sqrt{68.65^2+32.8^2}\approx 76.08

    Then use the sine rule again to solve for angle B.

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