Find Angle in Right Triangle Given Hyp and Opposite Side

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Discussion Overview

The discussion revolves around finding angles in right triangles given the lengths of the hypotenuse and the opposite side, as well as a related problem involving the calculation of the radius from the circumference of a circle. Participants explore the mathematical relationships and potential errors in calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant initially attempts to find an angle using the sine function but encounters an error due to the ratio exceeding 1.
  • Another participant points out that the sine function only accepts values between -1 and 1, suggesting the need to verify the triangle's dimensions.
  • A participant acknowledges their misunderstanding and retracts their earlier question after realizing the impossibility of the given triangle dimensions.
  • Several participants discuss the calculation of the radius from the circumference, with one participant asserting that dividing by 2π should yield the radius, but their calculation leads to an incorrect result.
  • Another participant emphasizes the importance of common sense in checking calculations, noting that dividing a smaller number by a larger number cannot yield a result greater than 1.
  • There is a correction regarding the arithmetic involved in the radius calculation, with a participant suggesting that the original poster may have made a basic error in understanding division.

Areas of Agreement / Disagreement

Participants generally agree on the mathematical principles involved, but there is disagreement regarding the specific calculations and interpretations of the problems presented. The discussion remains unresolved regarding the correct approach to the initial triangle problem.

Contextual Notes

Participants express uncertainty about the dimensions of the triangle and the calculations involved in determining the radius from the circumference. There are indications of missing assumptions and potential misunderstandings in basic arithmetic.

vysero
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If I have a right triangle and I know the hyp and the length of the opposite side of the angle I want then how do I find that angle? For instance: hyp = .9m and side opposite the angle I want = 1.5m. I tried dividing 1.5/.9 = 1.7 then I thought all i had to do was take the inverse sin of 1.7 and it would give me my angle but my calc gives me an error... So I am confused, how do I get the angle i want?
 
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vysero said:
If I have a right triangle and I know the hyp and the length of the opposite side of the angle I want then how do I find that angle? For instance: hyp = .9m and side opposite the angle I want = 1.5m. I tried dividing 1.5/.9 = 1.7 then I thought all i had to do was take the inverse sin of 1.7 and it would give me my angle but my calc gives me an error... So I am confused, how do I get the angle i want?

Sine oscillates between -1 and 1 so you can't take the inverse sine of 1.7, right? You should probably draw a picture and make sure you've got the relationships right.
 
I figured out that it's not going to work like that because I am asking for a value greater than 1... ok so I acted to brashly and asked for help before I thought my question through sorry guys ignore this post.
 
Anyway, the hypotenuse of a right triangle can't be never shorter than any cathetus, so you must have wrong lengths.
 
Okay so here is the problem. If I have a circumference of .9m and I want the radius of the circle then I just divide .9 by 2pi right? Which according to my calculator = 1.5 m... However, when I put this same problem into Microsoft mathematics it says r = .14 When I look at the solution steps it says something about how dividing by 2pi undoes the multiplication of 2pi... I guess I am missing something here.
 
adriaat said:
Anyway, the hypotenuse of a right triangle can't be never shorter than any cathetus, so you must have wrong lengths.

Plus 1 for cathetus. I've only ever heard the word "leg" in this context. Good knowledge! And welcome to the forum.

vysero said:
Okay so here is the problem. If I have a circumference of .9m and I want the radius of the circle then I just divide .9 by 2pi right? Which according to my calculator = 1.5 m...

That can't be right. One little math trick is to always compare calculator results to common sense. That's a lifesaver on tests. As you work, continually make common-sense estimates and see if your numbers are in the ballpark.

Pi is about 3, right? That's close enough for the moment. So what's 2pi? It's about 6. And if I take .9 and divide by 6, how can the answer be 1.5? If I start with .9, which is a little less than 1; and I divide it into 6 pieces; then each piece needs to be way less than 1. Can't be greater than 1. So right here if you're thinking about this as you go, you'd know you made a mistake.

I apologize for being a dinosaur here ... but you should put down that calculator. You're punching numbers in as a substitute for thinking about what's going on. Easy to do. But it can lead you astray. Better to just work this problem out on paper.

By the way, 9 divided by 6 is 1.5. You forgot the decimal point. But if you develop the habit of doing reality checks as you work, you'll avoid these kinds of errors on tests.
 
Last edited:
vysero said:
Okay so here is the problem. If I have a circumference of .9m and I want the radius of the circle then I just divide .9 by 2pi right? Which according to my calculator = 1.5 m...
Then get a new calculator or reread the manual! A small number divided by a larger number (2pi is larger than 6) cannot be larger than 1. .9 divided by [itex]2\pi[/itex] is about .14. Surely you aren't under the impression that [itex]\pi[/itex]= 0.314... but that would be my best guess- that you have drop a factor of 10.

However, when I put this same problem into Microsoft mathematics it says r = .14 When I look at the solution steps it says something about how dividing by 2pi undoes the multiplication of 2pi... I guess I am missing something here.
Uh, basic arithmetic? When you learn how to divide, in the second or third grade, you should learn that division is the reverse of multiplication.
 

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