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Posted by on Friday, February 10, 2012 at 6:18am.

the bases of a trapezoid are 22 and 12 respectively. The angles at the extremities of one base are 65 degree and 40 degree respectively find the two legs. Answer using law of sines pls

  • trigonometry - , Friday, February 10, 2012 at 8:17am

    Construct the trapezoid ABCD ,where AB || CD
    AB = 12 and CD = 22
    angle C=65 and angle D = 40

    Draw AE || BC where E is on CD
    So now ABCE is a parallelogram, and CE = 12
    which makes ED = 10

    Now look at triangle AED, by corresponding angles
    angle AED = 65°, angle D = 40 leaving angle DAE = 75°

    by sine law:
    AD/sin65 = 10/sin75
    AD = 10sin65/sin75 = 9.38

    by sine law:
    AE/sin40 = 10/sin75
    AE = 6.65

    but BC = AE, (||gram|

    So the side adjacent to the 65° angle is 6.65, the side adjacent to the 40° angle is 9.38

    check my arithmetic, I am only on my first coffee.

  • trigonometry - , Thursday, March 15, 2012 at 12:00am

    wow this really helps!! :D I have the same problem too haha thanks!!!

  • trigonometry - , Thursday, July 4, 2013 at 7:50am

    Hey thanks! I have the same problem. never thought it would be this easy.

  • trigonometry - , Friday, July 10, 2015 at 11:44pm

    tnx bro!

  • trigonometry - , Sunday, January 29, 2017 at 5:19pm

    where'd you get that 10 and 12

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