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JR Sauerland
Gold Member
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Seriously, I understand it sometimes, but when it hits me with something like this, I'm just baffled...
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 said: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 ?}##
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 ?}##BvU said:Multiply left and right with 50 ?, to get ##?^2 = 50 \times 32##
JR Sauerland said: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?
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 said:Do you understand how BvU went from
[tex]\frac{x}{50}=\frac{32}{x}[/tex]
to
[tex]x^2=50\times 32[/tex]
JR Sauerland said: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...
A similar triangle is a triangle that has the same shape as another triangle, but may have different sizes. This means that the corresponding angles of the two triangles are equal, and the corresponding sides are proportional.
To determine if two triangles are similar, you can use the AA (angle-angle) rule, which states that if two angles of one triangle are equal to two angles of another triangle, then the triangles are similar. You can also use the SAS (side-angle-side) or SSS (side-side-side) rule to determine similarity.
Similar triangles are important in geometry because they allow us to solve for unknown values in geometric figures. They also help us understand and prove geometric principles, such as the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides.
To solve problems involving similar triangles, you can use the proportionality theorem, which states that if two triangles are similar, then the ratio of any two corresponding sides is equal. You may also use the altitude-on-hypotenuse theorem or the midsegment theorem to solve for unknown values.
Yes, similar triangles can have different orientations. This means that they may be rotated, reflected, or translated, but their corresponding angles will still be equal and their corresponding sides will still be proportional, making them similar triangles.