Question 1117439
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 *[illustration CourseCorrectionCrop.jpg].

You have side-angle-side, so use the Law of Cosines:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t_1\ =\ \sqrt{b^2\ +\ t_2^2\ -\ 2bt_2\cos T_1}]


where *[tex \Large t_i] is the side opposite angle *[tex \Large T_i] and *[tex \Large b] is the side opposite angle *[tex \Large B]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t_1\ =\ \sqrt{20^2\ +\ 65^2\ -\ 2(20)(65)\cos 10^\circ}]


The rest is just calculator work.  You should examine the diagram to convince yourself that the answer is going to be larger than 45 but smaller than 65 as a means of checking the reasonableness of your calculator work.


Use the Law of Sines:


  
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{Sin A}{a}\ =\ \frac{Sin B}{b}\ =\ \frac{Sin C}{c}]


to solve the second problem in your submission.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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