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Math Help - Finding the measure of angles

  1. #1
    Junior Member Melancholy's Avatar
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    Finding the measure of angles

    Hello all,

    I joined because I have a quick question on some basic geometry/higher algebra 1. (and since you all seem to be college students, maybe you could help me out?)

    Here's my problem:

    The measures of two supplementary angles are 4r - 7 and r - 2. Find the measure of the angles.

    Now, from looking back at my notes, supplementary angles are angles that equal to 180 degrees. So... all that I've gathered from my info so far is that something having to do with 4r - 7 and r - 2 = 180. How do I set my problem up from there? (I can figure it out algebraicly when the equasion is set up)
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Melancholy View Post
    Hello all,

    I joined because I have a quick question on some basic geometry/higher algebra 1. (and since you all seem to be college students, maybe you could help me out?)

    Here's my problem:

    The measures of two supplementary angles are 4r - 7 and r - 2. Find the measure of the angles.

    Now, from looking back at my notes, supplementary angles are angles that equal to 180 degrees. So... all that I've gathered from my info so far is that something having to do with 4r - 7 and r - 2 = 180. How do I set my problem up from there? (I can figure it out algebraicly when the equasion is set up)
    solve (4r - 7) + (r - 2) = 180 for r. then plug your answer into the formula for each angle to find their measure
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  3. #3
    Junior Member Melancholy's Avatar
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    So then, the answers would be 37, and 147? Would it be in degrees?

    Another problem:

    Two angles are complementary. One angle measures 26 degrees more than the other. Find the measures of the angles.

    Erm... I know that one angle is 26 degrees more than the other!...... and I guess complementary angles have a sum of 90 degrees... so how do I set up the problem?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Melancholy View Post
    So then, the answers would be 37, and 147?
    no, the sum of those two angles is greater than 180 degrees, therefore, they are not supplementary as we were told. you did something wrong.

    Would it be in degrees?
    yes. we equated the angles to 180, that means we are working in degrees. if we were working in radians, we would equate them to \pi. besides, it would be clumsy to describe angles so large in radians, a circle has only 2 \pi radians, after all.

    Another problem:

    Two angles are complementary. One angle measures 26 degrees more than the other. Find the measures of the angles.

    Erm... I know that one angle is 26 degrees more than the other!...... and I guess complementary angles have a sum of 90 degrees... so how do I set up the problem?
    Let one of the angles be x
    Then the other angle is x + 26

    Since the angles are complimentary, we have:

    x + (x + 26) = 90

    Now conitnue
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  5. #5
    Junior Member Melancholy's Avatar
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    so I plugged in that value for and .

    My answer is and . I guess that is the answer... thank you!

    I think I found the answer to my first problem!! I believe I was doing something wrong in the algebra part...

    How I did it with algebra:








    Plugging in numbers:



    r - 2
    37.8 - 2
    35.8


    4r - 2
    4(37.8) - 7
    144.2


    The sum of those two numbers = 180... so I guess that means they are right!!... Thank you, Jhevon! You're a life saver!
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Melancholy View Post



    so I plugged in that value for and .

    My answer is and . I guess that is the answer... thank you!

    I think I found the answer to my first problem!! I believe I was doing something wrong in the algebra part...

    How I did it with algebra:








    Plugging in numbers:



    r - 2
    37.8 - 2
    35.8


    4r - 2
    4(37.8) - 7
    144.2


    The sum of those two numbers = 180... so I guess that means they are right!!... Thank you, Jhevon! You're a life saver!
    i didn't really do much, but ok, you're welcome. good job! and good luck with your class!
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  7. #7
    Junior Member Melancholy's Avatar
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    *princess voice*

    "Jhevon, oh Jehvon! Where arrrrrre yoouuuuu?!"

    Lol, question of the night:

    If m/ AGB = 4x + 7 and m / EGD = 71, find x.

    Find x. Find x. Okaaaaaay. *starts looking around for x...*

    The drawing they gave me looks somewhat like this:


    Oh yeah, an actual drawing from myself. I feel special.

    Please help
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Melancholy View Post
    *princess voice*

    "Jhevon, oh Jehvon! Where arrrrrre yoouuuuu?!"

    Lol, question of the night:

    If m/ AGB = 4x + 7 and m / EGD = 71, find x.
    \angle AGB and \angle EGD are what we call "opposite [or vertical] angles," these are angles that are opposite each other that are formed by the crossing of two straight lines.

    Theorem: Opposite angles are equal

    Thus, \angle AGB = \angle EGD

    \Rightarrow 4x + 7 = 71

    Now continue

    Find x. Find x. Okaaaaaay. *starts looking around for x...*
    this reminds me of a joke. see post #39 here. (it is the second to last post on the page, posted by Krizalid)


    The drawing they gave me looks somewhat like this:


    Oh yeah, an actual drawing from myself. I feel special.

    Please help
    nice drawing
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  9. #9
    Junior Member Melancholy's Avatar
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    Wow! A theorm even! Haven't gotten to those beauties yet. Can't wait lol.

    x = 16

    this reminds me of a joke. see post #39 here.
    Lol! How sarcastic! Sarcasm rules... that was hilarious!

    nice drawing
    Thank you for the compliment

    *sigh* I shall move on to my conjecture problems now. Might need some help with those soon. We'll see.
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Melancholy View Post
    Wow! A theorm even! Haven't gotten to those beauties yet. Can't wait lol.
    well, i think it's a theorem.

    if you're interested, you can look it up. doing a quick google search on "opposite angles" should give you all the information you need.

    x = 16
    correct

    Lol! How sarcastic! Sarcasm rules... that was hilarious!


    Thank you for the compliment

    *sigh* I shall move on to my conjecture problems now. Might need some help with those soon. We'll see.
    sure thing.

    i should have told you this before, but you should make sure to post new questions in new threads with the appropriate titles. it helps to keep things organized around here. up to now, all your questions have been about angles, so it wasn't that bad, but if the topics deviate vastly, you need a new thread
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  11. #11
    Junior Member Melancholy's Avatar
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    I'll remember that next time. I'm just used to the way my other forum works... they looove recylcing threads, so I apologize. I'll make a new thread with the right title next time.

    ... so repetative my post was... lol D. Do you guys white fontitize on here? Oh yeah... it's awesome...... nevermind....
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  12. #12
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Melancholy View Post
    ... so repetative my post was... lol D. Do you guys white fontitize on here? Oh yeah... it's awesome...... nevermind....
    yes we do, all the time

    you're going to get us in trouble with all this joking around, you know
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