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  1. #1
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    trig help

    B=5 deg C=125 deg b=200
    round to the nearest tenths and lengths
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  2. #2
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    Quote Originally Posted by SHORTY View Post
    B=5 deg C=125 deg b=200
    round to the nearest tenths and lengths
    And what exactly do you want us to do with this information? Ask us a question!

    -Dan
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  3. #3
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    Hello, Shorty!

    I assume the instructions are "Solve the triangle".

    You're expected to be familiar with the Law of SInes:
    . . \frac{a}{\sin A} = \frac{b}{\sin b} = \frac{c}{\sin C}



    B = 5^o,\;C = 125^o,\;b = 200
    Round to the nearest tenth.

    I prefer to organize the information in a chart.

    . . \begin{array}{ccc}A = \\B = \\ C = \end{array}<br />
\begin{array}{ccc}\boxed{\quad} \\ 5^o \\ 125^o\end{array}\quad<br />
\begin{array}{ccc}a = \\ b = \\ c = \end{array}<br />
\begin{array}{ccc}\boxed{\quad} \\ 200 \\ \boxed{\quad}\end{array}

    Since the angles total 180, we have: A \:= \:180^o - 5^o - 125^o \:= \:50^o
    Hence, we have:
    . . \begin{array}{ccc}A = \\B = \\ C = \end{array}<br />
\begin{array}{ccc}\boxed{50^o} \\ 5^o \\ 125^o\end{array}\quad<br />
\begin{array}{ccc}a = \\ b = \\ c = \end{array}<br />
\begin{array}{ccc}\boxed{\quad} \\ 200 \\ \boxed{\quad}\end{array}


    To find side c, use: . \frac{c}{\sin C} = \frac{b}{\sin B}\quad\Rightarrow\quad c \:=\:\frac{b\sin C}{\sin B}

    . . and we have: . c\:=\:\frac{200\sin125^o}{\sin5^o} \:\approx\:1879.7


    To find side a, use: . \frac{a}{\sin A} = \frac{b}{\sin B}\quad\Rightarrow\quad a \:=\:\frac{b\sin A}{\sin B}

    . . and we have: . a \:=\:\frac{200\sin50^o}{\sin5^o} \:\approx\:1757.9


    . . \boxed{\begin{array}{ccc}A = \\B = \\ C = \end{array}<br />
\begin{array}{ccc}\boxed{50^o} \\ 5^o \\ 125^o\end{array}\quad<br />
\begin{array}{ccc}a = \\ b = \\ c = \end{array}<br />
\begin{array}{ccc}\boxed{1757.9} \\ 200 \\ \boxed{1879.7}\end{array}} . . . There!

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