Anonymous
Anonymous asked in Education & ReferenceHomework Help · 10 years ago

Math circle geometry help. PIC INCLUDED! Double check my answers?

Hi!

I am doing this math project on "Using Chords to Determine Angles"

I'm not sure if I'm doing it right though. I know I have the idea, so don't tell me I shouldn't be doing it like that... I'm just not sure that the numbers are correct.

here is the assignment: http://www.flickr.com/photos/55227560@N08/54691316...

Here are my answers for the first three.

JLK: 42 deg

KJL: 90 deg

JNL: 25 deg

Am I doing it right?

... also I can't really figure out the last three...

Thanks!!

1 Answer

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  • Dragon
    Lv 6
    10 years ago
    Favorite Answer

    We know 1) <JKL = 25 deg (given), 2) <JLN = 42 deg (given), 3) Sum of the angles = 180 deg (proven elsewhere in your geometry book), and 4) the angles on each end of the base of an isosceles triangle are equal (proven elsewhere in your geometry book. 5) adjacent supplemental angles along a straight line equal 180 deg (proven elsewhere in your geometry book)

    So lets unravel the puzzle one angle at a time (except for maybe the last two)

    1) <JNL = 25 deg: As you concluded <JNL (also known as < KNL) = 25 deg since it is the other base angle of isosceles triangle KLN. (JKL being the other 25 deg angle which was one of the givens

    2) < KLJ = 88 deg Now we know <KLN = 180 - 50 deg or 130 deg and < JLK = 180 - 42 deg or 88 deg

    3) <KJL = 67 deg Re Triangle JKL; < KJL = 180 - 25 - 88 or 67 deg

    4) <NJL = 113 deg Since it joins the adjacent supplemental <LJK which = 67; or, if you prefer, since triangle JLN has 180 deg and we have defined <JNL as 25 deg and <JLN as 42 deg so <NJL = 180 - 25 - 42 = 113

    5) <NLM = 50 deg Since <KLN is 130 deg and is supplemental to < NLM (130 + 50 = 180)

    6) <NML = <MNL = 65 deg since we have an isosceles triangle LMN and the value of the angles is (180 - 50)/2 or 65 deg

    That's all folks! Isn't math fun?

    Source(s): Geometry textbook
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