# Math Help - Ambigious case for the law of sines?

1. ## Ambigious case for the law of sines?

Determine the number of solution for each possible triangle

B= 61 degrees a=12 b=8
The answer on the back of my book states zero.
Although I would think it has 2 because
12> b sin a when a is 29 degrees
Can anyone help me as I am slightly confused.

2. Originally Posted by homeylova223
Determine the number of solution for each possible triangle

B= 61 degrees a=12 b=8
The answer on the back of my book states zero.
Although I would think it has 2 because
12> b sin a when a is 29 degrees
Can anyone help me as I am slightly confused.
The triangle must obey the Sine Law, however..

$\displaystyle\frac{sin61^o}{8}\ \ne\ \frac{sin29^o}{12}$

$\displaystyle\frac{sinA}{a}=\frac{sinB}{b}\Rightar row\ sinA=\frac{12sin61^o}{8}=1.312$

which is not possible, since

$-1\ \le\ sinA\ \le\ 1$

3. Wikipedia offers and explanation and drawing which is helpful.

Law of sines - Wikipedia, the free encyclopedia

4. Hello, homeylova223!

Determine the number of solution for this triangle:
. . $B= 61^o,\;a=12,\; b=8$

The answer on the back of my book states: zero..

Code:
              C
*
/:\
/ : \
a=12 /  :  \ b=8
/   :   \
/    :    \
/     :
/61o   :
B * - - - + - - - * A
D

Draw altitude $\,CD$ to side $\,AB.$
In right triangle $CDB\!:\;\sin61^o \:=\:\dfrac{CD}{12} \quad\Rightarrow\quad CD \:=\:12\sin61^o \;\approx\;10.5$

Side $\,b$ is only 8 units long.

There is no triangle.