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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

 Jun 11, 2018
 #1
avatar+985 
+1

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

 Jun 11, 2018

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