I'm forgetting my geometry. Can I solve this triangle?

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The discussion revolves around solving a triangle for a physics problem, where the base is 56 meters, the height is 500 meters, and the difference between the two unknown sides is 4 meters. Participants confirm that the Pythagorean Theorem can be applied to find the sides, suggesting to split the triangle into two right triangles by dropping a perpendicular. The user initially struggles with how to split the base but eventually learns to set up equations based on the segments. Despite solving the geometry problem, the user realizes they approached the physics problem incorrectly. The conversation highlights the importance of geometry in solving physics-related problems.
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Homework Statement


So I'm doing a physics problem, and I think I'd get the right answer if I solved this triangle. I don't know any angles. The base is 56 meters long, its height is 500 meters, and the difference between the other two sides is 4 meters. Can I figure out the sides based on this information?
If not, can I at least find out the difference in position of the vertex between the two unknown sides if they were equal vs in this situation? I hope that makes sense.
 
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dzidziaud said:

Homework Statement


So I'm doing a physics problem, and I think I'd get the right answer if I solved this triangle. I don't know any angles. The base is 56 meters long, its height is 500 meters, and the difference between the other two sides is 4 meters. Can I figure out the sides based on this information?
If not, can I at least find out the difference in position of the vertex between the two unknown sides if they were equal vs in this situation? I hope that makes sense.
Yes. You have enough information to figure out the sides. Think Pythagorean Theorem.

Chet
 
Chestermiller said:
Yes. You have enough information to figure out the sides. Think Pythagorean Theorem.

Chet

I am assuming you mean by dropping a perpendicular that is the height of the triangle, so it is split into two right triangles? I can't figure out what to do from there though because I don't know how the 52 m side is split.
 
dzidziaud said:
I am assuming you mean by dropping a perpendicular that is the height of the triangle, so it is split into two right triangles? I can't figure out what to do from there though because I don't know how the 52 m side is split.
Split it into y and 56-y, and then get two equations in two unknowns. Of course, the other two sides are x and x+4.

Chet
 
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Thanks, I finally figured it out! Unfortunately, I got the wrong answer, so I'm approaching the actual physics problem wrong. At least I got some good geometry practice out of it :)
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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