# Find the Missing parst of the Triangle Using the Laws of Cosines

• Jan 21st 2014, 07:48 PM
Jarod_C
Find the Missing parst of the Triangle Using the Laws of Cosines
Hey people. Here's the question.

Find the missing parts of the triangle by using the laws of cosines.
What's known is, B=27, a=11, c=22

Solving for b,
$\displaystyle b^2=11^2+22^2-2(11)(22)(cosB)$
$\displaystyle b^2=121+484-2(242)(.891)$
$\displaystyle b^2=605-432$
$\displaystyle b^2=\sqrt{173}$
$\displaystyle b=13.2$

Solving for A
$\displaystyle \frac{11}{sinA}=\frac{13.2}{sin27}$

Cross multiply.
$\displaystyle \frac {sin27\cdot 11}{13.2}=.3783(sin^-^1)=~22$

Solving for C

$\displaystyle \frac{22\cdot sin27}{13.2}=.7567(sin^-^1)=~49$

180-27-49-22=82? Where did I go so wrong. This formula has always seemed to work for me.
• Jan 21st 2014, 07:59 PM
ibdutt
Re: Find the Missing parst of the Triangle Using the Laws of Cosines
But you have been asked tom use cosine formula and that it
cos A = [b^2 + c^2 - a^2 ]/ 2 bc
cos B = [c^2 + a^2 - b^2 ]/ 2 ac
cos C = [a^2 + b^2 - c^2 ]/ 2 ab
Use these expressions to get the result.
• Jan 21st 2014, 08:37 PM
Jarod_C
Re: Find the Missing parst of the Triangle Using the Laws of Cosines
Have you ever noticed there are 3 steps when using this forum. 1 post a question, 2 wait for answer 3 get answer then feel stupid. hahaha

Thanks, that solved it.