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Math Help - Checking work on triangles

  1. #1
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    Checking work on triangles

    Can someone check over my work?

    Lookouts in two fire towers, A and B located 10km apart sight a forest fire. Electronic equipment allows them to determine that the fire is at an angle of 71 from tower a and 100 from tower b. Which tower is closer to the fire and how far away is it.

    I drew a triangle with the information and did 71+100= 171-180=9 for angle C. So I got A=71, B=100, and C=9 then I used the A=10sin(100)/sin(9) and got 62.95 (rounded)

    2. On a baseball diamond with 90-ft sides, the pitcher's mound is 60.5 feet from home plate. How far is it from the pitcher's mound to third base?

    I made a picture of it and then found the following angles 45,90,45 and the hyp is 90, adj is 60.5 and then I did the a square + b square= c square and found 66.63 (rounded) did I do this right?

    3. Solve the equation 2sin x- (square root 3)- 0 for exact value of x in [0,2pie).

    I got pie/6, 5pie/6 for this did I do it right? I moved the square root of 3 over and then divided by the 2 and then found sin.
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  2. #2
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by goldenroll View Post
    Can someone check over my work?

    Lookouts in two fire towers, A and B located 10km apart sight a forest fire. Electronic equipment allows them to determine that the fire is at an angle of 71 from tower a and 100 from tower b. Which tower is closer to the fire and how far away is it.

    I drew a triangle with the information and did 71+100= 171-180=9 for angle C. So I got A=71, B=100, and C=9 then I used the A=10sin(100)/sin(9) and got 62.95 (rounded)

    2. On a baseball diamond with 90-ft sides, the pitcher's mound is 60.5 feet from home plate. How far is it from the pitcher's mound to third base?

    I made a picture of it and then found the following angles 45,90,45 and the hyp is 90, adj is 60.5 and then I did the a square + b square= c square and found 66.63 (rounded) did I do this right?

    3. Solve the equation 2sin x- (square root 3)- 0 for exact value of x in [0,2pie).

    I got pie/6, 5pie/6 for this did I do it right? I moved the square root of 3 over and then divided by the 2 and then found sin.
    On #1, you are correct in using the Law of Sines, but you have your values wrong. The angles are A = 71 B = 100 Fire (aka C) = 9 So using the Law of Sines:

    \frac{a}{sin(71)} = \frac{10}{sin(9)}

    On #2, what do you know about an isoceles triangle? If those two angles are equal, are not the sides opposite them also equal?

    For #3, I think you meant: 2sin(x) - \sqrt{3} = 0 right? It looks like - 0 in what you have. Nonetheless, your method is correct, but your answers are wrong. If you add \sqrt{3} to both sides and then divide by 2, you will have:
    sin(x) = \frac{\sqrt{3}}{2}. Now, where on your unit circle does the sinx =\frac{\sqrt{3}}{2} Your answers are where the COS(x) = \frac{\sqrt{3}}{2}
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  3. #3
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    Oh! so
    3. is pie/3, 2pie/3
    2. Wait and it can't be iso because then it wouldn't add up to the when doing a^2+b^2=c^2 and I already have the hyp and the adj numbers so all i have to do is a^2-c^2=b^2? or did I just go completely off track with this?
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  4. #4
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by goldenroll View Post
    Oh! so
    3. is pie/3, 2pie/3
    2. Wait and it can't be iso because then it wouldn't add up to the when doing a^2+b^2=c^2 and I already have the hyp and the adj numbers so all i have to do is a^2-c^2=b^2? or did I just go completely off track with this?
    #3 is right!!

    #2 The question doesn't really give you enough information. I mean is a pitcher's mound at the center of the baseball diamond? In other words, is it equidistant from each base? Because if it is NOT then the angle at the pitcher's mound will not be a 90 degree angle, and therefore you can not use the pathagorean theorem. But, I think you can assume that the line from the pitcher's mound to home base bisects the 90 degree angle at home base making it 45 degrees. See my point? But yes, if you know that there is a 90 deg angle and you have the other two sides you could use pathagorean theorem.
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