I'm having a lot of trouble with similar triangles

1. Jul 26, 2015

JR Sauerland

Seriously, I understand it sometimes, but when it hits me with something like this, I'm just baffled...

2. Jul 26, 2015

BvU

A matter of writing down $\cos\theta$ in two ways: for the left triangle it's ?/50 and for the right one it's 32/? So ${?\over 50} = {32\over ?}$

3. Jul 26, 2015

JR Sauerland

See, that's the part I get because of SOH CAH TOA. It's cosine, CAH, Adjacent over the Hypotenuse. The other one is completely the opposite. I set the problem up just like you did, but I really don't know where to go from there. ${?\over 50} = {32\over ?}$ That is what is baffling me.

4. Jul 26, 2015

BvU

Multiply left and right with 50 ?, to get $?^2 = 50 \times 32$

5. Jul 26, 2015

JR Sauerland

Perfect! Now I completely understand the logic behind it! On another note, ${30\over ?} = {?\over 32}$, if it had been formatted like this, do I simply take the reciprocal like this: ${?\over 30} = {32\over ?}$
Basically, I'm getting at the fact that the variable (which typically is x) isn't supposed to be on the bottom?

6. Jul 26, 2015

Mentallic

Do you understand how BvU went from

$$\frac{x}{50}=\frac{32}{x}$$
to
$$x^2=50\times 32$$

7. Jul 26, 2015

JR Sauerland

I think so. Didn't he multiply both sides by (50x)? It eliminated x in the denominator of the right side, and squared the x in the left side due to the fact that it was on top. I've heard you could cross multiple too though, although I don't know if it would be completely different or not...

8. Jul 26, 2015

Mentallic

Right, so apply the same idea to your new problem. Taking the reciprocal of both sides is a valid operation, but it doesn't help you because you want to remove the variable from the denominator, which just so happens to give us a quadratic (squared value of x) in this sort of problem.

Also, cross multiplying is exactly multiplying both sides by 50x. It's just a means to help students visualize or to rote learn the process more easily.