+83 1)Find BC.
Picture: https://latex.artofproblemsolving.com/f/e/4/fe45662f573f87e30f9f30d0b1ef1f94b4c22fc2.png
2)Find AC.
Picture: https://latex.artofproblemsolving.com/6/8/d/68dd918aac5d85a0b2a5b3b9c6a79eabd3669f01.png
3)In triangle ABC, CA = 4sqrt2, CB = 4sqrt3, and angle A = 60 degrees. What is B in degrees?
4)Find AC.
Picture: https://latex.artofproblemsolving.com/9/1/d/91d520533f0fb8cb325932013b2aebf0db9bde54.png
5)Find sin B.
Picture:https://latex.artofproblemsolving.com/4/7/a/47a312412eff41c09e4ce21e5edbb7e52e8b151c.png
6)Find cos A.
Picture: https://latex.artofproblemsolving.com/6/4/a/64ae47768bae6b24ab792c44a92c5de244955cad.png
7)In triangle ABC, sin A = 4/5. Find cos B.
Picture: https://latex.artofproblemsolving.com/c/8/3/c83240e209d830cd2275ed9f771185a1d1c17733.png
8)Find DB in the diagram below.
Picture: https://latex.artofproblemsolving.com/f/6/f/f6f917a5c86b363a5cdf8871300d061dcd54fc0b.png
+983 Hey FiestyGeco!
I can only help you with 3, since the pictures won't show.
For 3, we use \(c^2=a^2+b^2-2ab\cos\gamma\), the law of cosines.
This formula generalizes the Pythagorean Theorem, since \(\cos90º=0\Rightarrow a^2+b^2=c^2\)
We can write: \(CB^2=CA^2+AB^2-2(AC)(AB)-\cos{A}\)
\((4\sqrt3)^2=(4\sqrt2)^2+AB^2-2(4\sqrt2)(AB)-\cos60\\ 48=32+AB^2-8\sqrt2AB-\frac12\\ \)
After solving for AB, you use the same formula to solve for the angle.
This doesn't look like the best method, does anyone know a less rigid way of doing this?
I hope this helped,
Gavin
