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The ambigous case

  1. Apr 3, 2006 #1


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    I dont understand how to do these questions, involving angles less than 90 degrees. I know the "rules" you have to follow like if a<bsina there are two triangles etc. but what i dont understand is when there is a question, with no picture, how are you supposed to know witch one is the "a" and witch one it the "b".

    For example
    Determine the number of possible triangles that could be drawn with the given measurments

    trianglePQR, where..

    c=30 cm

    Pleas help!
  2. jcsd
  3. Apr 3, 2006 #2


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    Staff Emeritus
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    Surely the convention was explained to you? "C" is the angle opposite side "c", "B" is the angle opposite side "b", and "A" is the angle opposite side "a".

    Given "angle B= 27 degrees, b= 25 cm., c= 30 cm., how would you draw the triangle? Here's what I would do. Draw a horizontal line and label it "c". Make it 30 cm long. Since angle C is opposite that, angle B and angle A are at the ends of that line. Choose whichever end you prefer, label it "B" and draw measure off a 27 degree angle. Of course, side "b" is opposite that so the side you just drawn is "a" and you don't know how long it will be- just extend it as long as you can. Now go to the other end of your horizontal line. That's, since its the only possibility left, angle "A" and the third side there is side "b". You know it's 25 cm long but you don't know the angle. That's what compasses are for- strike an arc of a circle with radius 25 cm. Where it crosses side "a" is the third vertex of the triangle- angle "C". There are 3 possiblities: that arc might cut side "a" twice, it might just touch it once, or it might no touch it at all. In the second possibility, just touching, side "a" would be tangent to the circle and angle "C" would have to be a right angle. That would be the case if side "b" were equal to length of c times sin(B)= 30 sin(27). If b is longer than that, there are two triangles, less than that, none.
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