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

• Dec 4th 2006, 12:38 PM
kcsteven
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
• Dec 4th 2006, 02:41 PM
Soroban
Hello, Keith!

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

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

Quote:

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.

. . $\begin{array}{ccc}A\;= \\ B\;= \\ C\;= \end{array}
\begin{array}{ccc} a\:= \\ b\:= \\ c\:=\end{array}
\begin{array}{ccc}5 \\ 3 \\ \boxed{\quad}\end{array}$

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:

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

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:

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

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

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

Our chart looks like this:

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

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:

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

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

**

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.