#2**+5 **

I have been asked to finish this one - I am your humble servant (only this time lol) so I shall be pleased to do so.

$$\begin{array}{rlll}

\frac{12}{sinB}&=&\frac{11}{sin30}\\\\

\frac{sinB}{12}&=&\frac{sin30}{11}\\\\

SinB&=&\frac{12sin30}{11}\\\\

B&=&asin\left( {\frac{12sin30}{11}}\right)\\\\

B&=&33.056^0\qquad &\mbox{First quadrant solution correct to 3 dec places} \\\\

B&=&180-33.056=146.944\qquad &\mbox{Second quadrant solution correct to 3 dec places}\\\\

\end{array}$$

$$If B is $33^0$ then \\

$C=180-30-33=117^0$\\\\

If B is $147^0$ then \\

$C=180-30-147=3^0$\\\\

So the 2 possible solutions are $

Melody Sep 1, 2014

#1**+5 **

I've have constructed the triangle to show you that there are exactly 2 solutions solutions.

This triangle is drawn to scale.

We need to find angle B first.

You can do this using the sine rule but remember, since sin is positive in the first and second quadrant there will be an accute angle B and an obtuse angle B - Two possible solutions for B

Once you have angle B you can find Angle C by using angle sum of a triangle. Once agian there will be 2 solutions.

You try finding this these angles.

If you get stuck again put another post after this one to ask for more help.

Melody Sep 1, 2014

#2**+5 **

Best Answer

I have been asked to finish this one - I am your humble servant (only this time lol) so I shall be pleased to do so.

$$\begin{array}{rlll}

\frac{12}{sinB}&=&\frac{11}{sin30}\\\\

\frac{sinB}{12}&=&\frac{sin30}{11}\\\\

SinB&=&\frac{12sin30}{11}\\\\

B&=&asin\left( {\frac{12sin30}{11}}\right)\\\\

B&=&33.056^0\qquad &\mbox{First quadrant solution correct to 3 dec places} \\\\

B&=&180-33.056=146.944\qquad &\mbox{Second quadrant solution correct to 3 dec places}\\\\

\end{array}$$

$$If B is $33^0$ then \\

$C=180-30-33=117^0$\\\\

If B is $147^0$ then \\

$C=180-30-147=3^0$\\\\

So the 2 possible solutions are $

Melody Sep 1, 2014