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 enter image description here

and a divide example enter image description here


  • $\begingroup$ 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. $\endgroup$ – André Nicolas Jan 22 '13 at 4:10
  • $\begingroup$ @AndréNicolas They only want me to find one side, the side with the "?". $\endgroup$ – Tyler Zika Jan 22 '13 at 4:12
  • $\begingroup$ Are they telling you the sine, or cosine, or tangent of any of the angles? $\endgroup$ – André Nicolas Jan 22 '13 at 4:16
  • $\begingroup$ @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. $\endgroup$ – Tyler Zika Jan 22 '13 at 4:24
  • $\begingroup$ @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. $\endgroup$ – 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 $3$.


  • $\begingroup$ Thank you for breaking it down! $\endgroup$ – Tyler Zika Jan 22 '13 at 9:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.