# Ambigious case for the law of sines?

• Dec 25th 2010, 01:27 PM
homeylova223
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.
• Dec 25th 2010, 01:43 PM
Quote:

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

$-1\ \le\ sinA\ \le\ 1$
• Dec 25th 2010, 01:44 PM
dwsmith
Wikipedia offers and explanation and drawing which is helpful.

Law of sines - Wikipedia, the free encyclopedia
• Dec 25th 2010, 08:09 PM
Soroban
Hello, homeylova223!

Quote:

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..

              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.