I'm having a lot of trouble with similar triangles

[JR Sauerland]

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

 

[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 ${?\over 50} = {32\over ?}$

 

[JR Sauerland]

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 ${?\over 50} = {32\over ?}$

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.

 

[BvU]

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

 

[JR Sauerland]

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

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?

 

[Mentallic]

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?

Do you understand how BvU went from

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

 

[JR Sauerland]

Do you understand how BvU went from

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

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...

 

[Mentallic]

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...

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.