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Math Help - solve the triangle, but this is not a right triangle

  1. #1
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    solve the triangle, but this is not a right triangle

    This is one of the problems on the review sheet for our next test.
    solve the triangle (not a right triangle)
    given B=20 degrees, a= 5, b=3
    she told us that this one has two triangles (34.8, 125.2, 7.2), (145.2, 34.8, 5)
    I am not sure how to tackle this one.
    Thank you!
    Keith Stevens
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  2. #2
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    Hello, Keith!

    This problem requires the Law of Sines.
    There are two forms, but one is the reciprocal of the other.

    . . \frac{a}{\sin A} \,=\,\frac{b}{\sin B} \,=\,\frac{c}{\sin C}\qquad\quad\frac{\sin A}{a}\,=\,\frac{\sin B}{b}\,=\,\frac{\sin C}{c}


    Solve the triangle: . B = 20^o,\;a = 5,\;b=3

    She told us that this one has two triangles: .
    . . . the dreaded Ambiguous Case!
    . . (34.8, 125.2, 7.2), (145.2, 34.8, 5)
    . . . Her second one is wrong!

    I like to organize the facts in a chart.

    . . \begin{array}{ccc}A\;= \\ B\;= \\ C\;= \end{array}<br />
\begin{array}{ccc}\boxed{\quad} \\ 20^o \\ \boxed{\quad} \end{array}\qquad<br />
\begin{array}{ccc} a\:= \\ b\:= \\ c\:=\end{array}<br />
\begin{array}{ccc}5 \\ 3 \\ \boxed{\quad}\end{array}


    We can find angle A with: . \frac{\sin A}{a} \,=\,\frac{\sin B}{b}\quad\Rightarrow\quad \sin A \,=\,\frac{a\sin B}{b}

    We have: . \sin A \:=\:\frac{5\sin20^o}{3} \:=\:0.570033572

    Hence: .  A \:=\:34.8^o or 145.2^o **

    We will solve the triangle for both values of A.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    [1] A = 34.8^o\quad\Rightarrow\quad C \:=\:180^o - 34.8^o - 20^o \:=\:125.2^o

    Our chart looks like this:

    . . \begin{array}{ccc}A\;= \\ B\;= \\ C\;= \end{array}<br />
\begin{array}{ccc}34.85^o \\ 20^o \\ 125.2^o \end{array}\qquad<br />
\begin{array}{ccc} a\:= \\ b\:= \\ c\:=\end{array}<br />
\begin{array}{ccc}5 \\ 3 \\ \boxed{\quad}\end{array}

    We need only side c.
    We will use: . \frac{c}{\sin C} = \frac{b}{\sin B}\quad\Rightarrow\quad c \:=\:\frac{b\sin C}{\sin B}

    We have: . c \:=\:\frac{3\sin125.2^o}{\sin20^o} \:=\:7.167515548

    Therefore: . c \:\approx\:7.2


    And this our completed chart:

    . . \boxed{\begin{array}{ccc}A\;= \\ B\;= \\ C\;= \end{array}<br />
\begin{array}{ccc}34.8^o \\ 20^o \\ 125.2^o \end{array}\qquad<br />
\begin{array}{ccc} a\:= \\ b\:= \\ c\:=\end{array}<br />
\begin{array}{ccc}5 \\ 3 \\ 7.2\end{array}}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    [2] A = 145.2^o\quad\Rightarrow\quad C \:=\:180^o - 145.2^o - 20^o \:=\:14.8^o

    Our chart looks like this:

    . . \begin{array}{ccc}A\;= \\ B\;= \\ C\;= \end{array}<br />
\begin{array}{ccc}145.2^o \\ 20^o \\ 14.8^o \end{array}\qquad<br />
\begin{array}{ccc} a\:= \\ b\:= \\ c\:=\end{array}<br />
\begin{array}{ccc}5 \\ 3 \\ \boxed{\quad}\end{array}

    We need only side c.
    We will use: . \frac{c}{\sin C} = \frac{b}{\sin B}\quad\Rightarrow\quad c \:=\:\frac{b\sin C}{\sin B}

    We have: . c \:=\:\frac{3\sin14.8^o}{\sin20^o} \:=\:2.24062-293

    Therefore: . c \:\approx\:2.2


    And this our completed chart:

    . . \boxed{\begin{array}{ccc}A\;= \\ B\;= \\ C\;= \end{array}<br />
\begin{array}{ccc}145.2^o \\ 20^o \\ 14.8^o \end{array}\qquad<br />
\begin{array}{ccc} a\:= \\ b\:= \\ c\:=\end{array}<br />
\begin{array}{ccc}5 \\ 3 \\ 2.2\end{array}}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **

    Whenever we use \boxed{\sin^{-1}}, there are two possible answers.

    Our calculator gives us only the acute angle (in Quadrant 1).
    . . The other angle is its supplement (subtract from 180).
    We must be constantly alert to the possibility of a second angle.

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