# Thread: solve the triangle, but this is not a right triangle

1. ## 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

2. 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}
\begin{array}{ccc} a\:= \\ b\:= \\ c\:=\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}
\begin{array}{ccc}34.85^o \\ 20^o \\ 125.2^o \end{array}\qquad
\begin{array}{ccc} a\:= \\ b\:= \\ c\:=\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:

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

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

[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}
\begin{array}{ccc}145.2^o \\ 20^o \\ 14.8^o \end{array}\qquad
\begin{array}{ccc} a\:= \\ b\:= \\ c\:=\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:

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

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

**

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.