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Math Help - Solving right triangle

  1. #1
    Junior Member
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    Solving right triangle

    I am in need of help solving this problem. I got the missing side as sqrt (34), but that all I've got.

    And this is kind of separate from this problem, but I don't really understand the Pythagorean Theorem with these problems. This one made sense, but another one I did had 40 as the hypotenus, and 41 as the opposite. I had to find "c". Well, I ended up with a=sqrt(40^2-41^2), because that makes sense, right? It turned out to be 41^2-40^2, but how? Wouldn't that have been sqrt(-41^2+40^2) if written the other way? I took a^2+b^2=c^2 and solved for "a". Is this not correct? I'm very confused, and any help would be appreciated! But the main focus of this thread is in my attachment. I didn't know how to write those characters so I drew up the problem.

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  2. #2
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    Since you already got all the sides, all you need to do is plug in the numbers.

    Sin = opposite/hypotenuse
    Cos = adjacent/hypotenuse
    Tan = opposite/adjacent

    In other words "SOHCAHTOA."

    Cot = inverse of Tan = opposite/adjacent
    Sec = inverse of Cos = hypotenuse/adjacent
    Csc = inverse of Sin = hypotenuse/opposite

    Using this information, sin fishy (alpha?) would be 3/radical 34. In turn, cos B thingy would also be 3/radical 34.

    Sorry, I forget the names of the thingies. Haha. I hope this helps.

    Edit: For your question thingy about the Theorem, are you sure the read the problem correctly? According to the answer, 41 should be the hypotenuse.
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Detroit, MI
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    Quote Originally Posted by Chinnie15 View Post
    I am in need of help solving this problem. I got the missing side as sqrt (34), but that all I've got.

    And this is kind of separate from this problem, but I don't really understand the Pythagorean Theorem with these problems. This one made sense, but another one I did had 40 as the hypotenus, and 41 as the opposite. I had to find "c". Well, I ended up with a=sqrt(40^2-41^2), because that makes sense, right? It turned out to be 41^2-40^2, but how? Wouldn't that have been sqrt(-41^2+40^2) if written the other way? I took a^2+b^2=c^2 and solved for "a". Is this not correct? I'm very confused, and any help would be appreciated! But the main focus of this thread is in my attachment. I didn't know how to write those characters so I drew up the problem.

    Begin by using the pythagorean theorem to find the hypotenuse. Then use the definitions of the trigonometric functions to find the indicated quantities.
    Do you know these definitions?
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  4. #4
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    Thanks! I know the definitions, but I'm having a hard time applying them and how to know which to use and when. Oh, I found my mistake on the one problem. It was the hypotenuse. I hate triangles, lol. They all look the same to me.

    Thank you again!
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