# Why do you multiply one way and divide the other way with these trig problems?

I am practicing finding a side of an angle on Khan Academy. I understand SOH CAH TOA and which sin, cos, tan to choose from. But, I don't understand why they multiply sometimes to find the side and divide other times. I am using a calculator.

Here is a multiply example

and a divide example

thanks.

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It is not clear in either case what the problem is. In the first picture, the hypotenuse is $\sqrt{97}$. Is there any additional information? If there is not, we cannot determine the sides. – André Nicolas Jan 22 '13 at 4:10
@AndréNicolas They only want me to find one side, the side with the "?". – Tyler Zika Jan 22 '13 at 4:12
Are they telling you the sine, or cosine, or tangent of any of the angles? – André Nicolas Jan 22 '13 at 4:16
@AndréNicolas they are telling me the sin, cos, and tan of just the sides I believe. From that information, they want me to figure out just one side. – Tyler Zika Jan 22 '13 at 4:24
@AndréNicolas at the end of each screen shot that I am showing, they show the steps on how to find the side. I don't understand the logic behind this. – Tyler Zika Jan 22 '13 at 4:25

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 $3$.