In the 1st problem, I think you understand it as far as the next-to-last line, $$\sin A={BC\over\sqrt{97}}$$ Also, you are given the value $$\sin A={9\sqrt{97}\over97}$$ So you have $${9\sqrt{97}\over97}={BC\over\sqrt{97}}$$ Since you want $BC$, you multiply both sides of this equation by $\sqrt{97}$ --- that gets $BC$ all by itself on one side of the equation, which is what you want. 

In the second problem, you understand $$\tan A={10\over AC}$$ You want to isolate $AC$. This can be done by multiplying both sides by $AC$, getting $$(AC)(\tan A)=10$$ and then dividing both sides by $\tan A$ to get $$AC={10\over\tan A}$$

Now you are given $$\tan A={10\over3}$$ so you have $$AC={10\over10/3}$$ which simplifies to $3$. 

OK?

Why do you multiply one way and divide the other way with these trig problems?
If I add entities to a moc and I execute a fetch request on that moc before its document gets a chance to save, will my fetch contain the newly added entities? (I'm unsure how to test this out because of the auto-save feature)
