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Math Help - right triangle trig

  1. #1
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    right triangle trig

    ok so in this problem..
    i am told to solve for all sides and angles in the right triangle
    so the bottom side of the triangle is 6.2 in length
    and all the other sides are unknown
    in angles the the triangle is a right triangle and that angle is located on the left bottom side of the triangle, on the lower right angle the angle is 25 degrees

    what are the missing sides and lengths?
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  2. #2
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    Find your vague description solve the missing lengths \cos 25^{\circ} = \frac{6.2}{x} and \sin 25^{\circ} = \frac{y}{6.2}
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  3. #3
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    x=6.81
    y=14.66
    ?
    or y= 0.06
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  4. #4
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    Quote Originally Posted by gabriel View Post
    ok so in this problem..
    i am told to solve for all sides and angles in the right triangle
    so the bottom side of the triangle is 6.2 in length
    and all the other sides are unknown
    in angles the the triangle is a right triangle and that angle is located on the left bottom side of the triangle, on the lower right angle the angle is 25 degrees

    what are the missing sides and lengths?
    make a sketch, then set up a couple of trig ratios ...

    \displaystyle \tan(25^\circ) = \frac{a}{6.2}

    \displaystyle \cos(25^\circ) = \frac{6.2}{b}

    solve for a and b
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  5. #5
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    when i solve for "a" I divide tan(25) by 6.2 and i get approximately 0.0752
    when i solve for "b" I multiply cos(25) by 6.2 and i get approximately 5.62
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  6. #6
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    Quote Originally Posted by gabriel View Post
    when i solve for "a" I divide tan(25) by 6.2 and i get approximately 0.0752
    when i solve for "b" I multiply cos(25) by 6.2 and i get approximately 5.62
    From skeeter's diagram

    tan25^o=\displaystyle\huge\frac{a}{6.2}

    If both sides are equal, then 6.2 times both sides are also equal,
    just as in the following..

    \frac{6}{2}=3\ \Rightarrow\ 2\frac{6}{2}=2(3)\ \Rightarrow\ \frac{2}{2}6=6

    Therefore

    (6.2)tan25^o=6.2\frac{a}{6.2}=\frac{6.2}{6.2}a=(1)  a=a


    cos25^o=\displaystyle\huge\frac{6.2}{b}\ \Rightarrow\ \frac{cos25^o}{1}=\frac{6.2}{b}

    Invert that..

    \Rightarrow\ \displaystyle\huge\frac{1}{cos25^o}=\frac{b}{6.2}

    Now multiply both sides by 6.2 to be able to calculate b.
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  7. #7
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    cos(25)*6.2 = approximately 5.62
    b=5.62
    and (6.2)tan25 =2.89
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  8. #8
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    Quote Originally Posted by gabriel View Post
    cos(25)*6.2 = approximately 5.62
    b=5.62
    and (6.2)tan25 =2.89
    Have a look at skeeter's diagram again..

    is b>6.2 or is b<6.2 ?

    What happens to \frac{1}{cos25^o}=\frac{b}{6.2}

    when you multiply both sides by 6.2?

    6.2\left(\frac{1}{cos25^o}\right)=6.2\left(\frac{b  }{6.2}\right)

    What is the next line?
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  9. #9
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    6.2(1/cos(25) = 6.84 = b
    and 6.2(6.84/6.82)=6.84
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  10. #10
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    Quote Originally Posted by gabriel View Post
    6.2(1/cos(25)) = 6.84 = b
    and 6.2(6.84/6.2)=6.84
    Yes,
    but the important thing is for you to be able to think it through...

    cos25^o=\frac{6.2}{b}

    We want to find out what b is, using cos25^o and 6.2.

    We need b on one side and the other two combined correctly to calculate b.
    There are lots of ways to combine two numbers..add, subtract, multiply, divide, etc.
    but there is only one way to combine those numbers to get b.

    b being in the denominator position is inconvenient,
    so first multiply both sides by b.

    cos25^o=\frac{6.2}{b}\ \Rightarrow\ bcos25^o=b\left(\frac{6.2}{b}\right)=\frac{b6.2}{b  }=\frac{b}{b}6.2=(1)6.2=6.2

    That gets b "above the line.
    However, it's now multiplied by cos25^o,

    therefore if we divide both sides by cos25^o, we are left with b all alone on the left,
    which is what we wanted...

    bcos25^o=6.2\ \Rightarrow\ \frac{bcos25^o}{cos25^o}=\frac{6.2}{cos25^o}\ \Rightarrow\ b(1)=b=\frac{6.2}{cos25^o}

    The important thing is to understand why you take the steps,
    not just "do this", then "do that"...

    Or understand the way shown earlier by inverting the fractions.
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  11. #11
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    i understand know but is it the same when you involve in tangents such as side "a" where u have to apply tan(25) to find side a?
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  12. #12
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    Quote Originally Posted by gabriel View Post
    i understand know but is it the same when you involve in tangents such as side "a" where u have to apply tan(25) to find side a?
    It all depends on where the side you are looking for is in the fraction,
    so you may have to do a slightly different calculation,
    but the idea is to get the side you want "by itself".

    In this case....

    tan25^o=\frac{side\ opposite\ 25^o}{side\ adjacent\ 25^o}=\frac{a}{6.2}

    This time we can get "a" by itself in one step by multiplying both sides by 6.2
    So it's faster!

    Finding "b" takes a little longer, because we begin with it under the line.
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  13. #13
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    so side "a" would be = to 2.89? if tan(25)*6.2=2.89?
    the fraction should be a/6.2?
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  14. #14
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    Quote Originally Posted by gabriel View Post
    so side "a" would be = to 2.89? if tan(25)*6.2=2.89?
    the fraction should be a/6.2?
    Yes,

    beginning with the fraction tan25^o=\frac{a}{6.2}

    we only need to multiply both sides by 6.2 to get (6.2)tan25^o=a

    Last step is to simply multiply whatever the calculator gives you for tan25 by 6.2.
    That's how you figure out "a" without having to draw the triangle and measure lengths.
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  15. #15
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    ok just checking
    and one more question if i could
    if u look back @ skeeter.s drawing, how do i find the missing angle @ the top?
    would it be 90+25=115 and 180-115=65 degrees?
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