# Why do I have to flip the equation in this problem?

1. Jul 27, 2015

### JR Sauerland

Another similar triangles problem... Woo hoo. This one was way more bizarre than any other, and is exactly why I have difficulty with these problems. For some reason, the problem expects me to have some inherent intuition and automatically know I'm supposed to flip the problem randomly, and gives absolutely no explanation as to why. Cosine (CAH) is Adjacent over Hypotenuse. For whatever reason, they flipped it, making it Hypotenuse over Adjacent, and didn't explain in the steps why. This is why I am having a hard time learning similar triangles. There's obviously a hidden step I'm missing because it isn't being explained, just done.

2. Jul 27, 2015

### DEvens

Do you mean that when they said they want the length of B'C, and so solved for B'C in the equation, you did not understand that?

3. Jul 27, 2015

### JR Sauerland

Sure, I guess that is better than anything… They didn't give any description of what they were doing at all, so it's pretty unclear to me what's going on

4. Jul 27, 2015

### Mentallic

But they ended with B'D = ... which would imply that their intent was to find the length of B'D. If you were asked to solve for B'C in the equation

$$\frac{DC}{B'C}=\cos\left(∠B'CD\right)$$

Given that you know the other values, how would you have done it?

5. Jul 27, 2015

### JR Sauerland

In the question, it actually did not even state that BC is the same length as BC or AD. The only time it actually stated that was after I opted to show the answer and solution. I suppose it's one of those things where you're inherently supposed to know. Just like in statistics where they ask you the chance of joined a king or a jack and a standard deck of 32 cards, they never tell you how many jacks or kings are in a standard deck of 52 cards to begin with. I don't play poker, and I also don't fold papers in half or diagonally on my spare time so I wouldn't know that.

But to answer your question, I had thought I was supposed to create an equation of opposite over adjacent, and take tangent of the angle to get B'D, and then from there I was lost.

6. Jul 27, 2015

### JR Sauerland

Additionally, instead of just randomly flipping the equation with no explanation, why couldn't they just phrase it a little differently? 1/COS(x) is the secant. If they would've literally just said to take the secant, which is hypotenuse over adjacent... Idk. I'm just not used to this type of problem I guess.

7. Jul 27, 2015

### Mentallic

I don't either, but it makes sense that if you fold the corner B over to B' and crease the paper along the line stated, then clearly BC = B'C because you've just folded BC over to B'C.

That's also a valid method of solving for B'D, but it doesn't help with finding B'C (which we are doing so that we can find the dimensions of the paper since B'C = BC). The solution also opted to use the Pythagorean theorem to find B'D as opposed to using tan as you would have done, likely just to remind you that you can use it as it often goes hand-in-hand with trigonometry.

It's not randomly flipping. It's called taking the reciprocal and it's a valid operation.

If $x=y$ then I'm sure you'll agree that $\frac{1}{x}=\frac{1}{y}$

Also,
$$\frac{1}{\left(\frac{a}{b}\right)}=\frac{b}{a}$$
so the reciprocal is essentially flipping the fraction. Hence, if you have

$$\frac{a}{b}=\frac{c}{d}$$
then we can take the reciprocal of both sides to get
$$\frac{1}{\left(\frac{a}{b}\right)}=\frac{1}{\left(\frac{c}{d}\right)}$$
$$\frac{b}{a}=\frac{d}{c}$$

But also multiplying both sides by the lowest common denominator which in this case is bd would also be a way of solving for a value that's found in the denominator. Cross multiplying is the quick way to think about this operation, just as how flipping the fraction is the quick way of thinking about taking the reciprocal.

Going from
$$\sec{x}=\frac{a}{b}$$
straight to
$$\cos{x}=\frac{b}{a}$$
would've likely caused confusion for other students too.

8. Jul 27, 2015

### SteamKing

Staff Emeritus
It would also help us if you would post the complete problem statement. As it stands, the image in the OP appears to start in the middle of the solution.