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# in triangle ABC if AC is 12 BC is 11 and angle A is 30 angle C could be what?

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In ΔABC, if AC= 12, BC = 11 and m<A= 30, angle C could be...

Guest Sep 1, 2014

#2
+91469
+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.

$$\begin{array}{rlll} \frac{12}{sinB}&=&\frac{11}{sin30}\\\\ \frac{sinB}{12}&=&\frac{sin30}{11}\\\\ SinB&=&\frac{12sin30}{11}\\\\ B&=&asin\left( {\frac{12sin30}{11}}\right)\\\\ B&=&33.056^0\qquad &\mbox{First quadrant solution correct to 3 dec places} \\\\ B&=&180-33.056=146.944\qquad &\mbox{Second quadrant solution correct to 3 dec places}\\\\ \end{array}$$

$$If B is 33^0 then \\ C=180-30-33=117^0\\\\ If B is 147^0 then \\ C=180-30-147=3^0\\\\ So the 2 possible solutions are  Melody Sep 1, 2014 Sort: ### 3+0 Answers #1 +91469 +5 I've have constructed the triangle to show you that there are exactly 2 solutions solutions. This triangle is drawn to scale. We need to find angle B first. You can do this using the sine rule but remember, since sin is positive in the first and second quadrant there will be an accute angle B and an obtuse angle B - Two possible solutions for B Once you have angle B you can find Angle C by using angle sum of a triangle. Once agian there will be 2 solutions. You try finding this these angles. If you get stuck again put another post after this one to ask for more help. Melody Sep 1, 2014 #2 +91469 +5 Best Answer I have been asked to finish this one - I am your humble servant (only this time lol) so I shall be pleased to do so.$$\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)\\\\
\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
#3
+81032
0

Thanks, Melody, for that one !!!

CPhill  Sep 1, 2014

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