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Advanced trigonometry

  1. May 6, 2007 #1
    I found the area of this triangle to be 56.4 (to 3sf) easily but i can't work out the length of CX. any ideas?

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  3. May 6, 2007 #2


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    Did you try the cosine law?

    c2= a2+ b2- 2ab cos(C)

    a= 15, b= 8, and C= 70 degrees.
  4. May 6, 2007 #3
    sorry i think i worded my question incrorrectly. I'm looking for the length BX. Angle CXB is a right angle. I can't use the cosine rule because i only know one length (15) and one angle (90degrees)
  5. May 6, 2007 #4


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    In other words, x is the point at the foot of the altitude! I was thinking x was th length of AB. You have two right triangles, CXA and CXB. The two given lengths are the lengths of the hypotenuses. Let the two angles at C be a and b. You know that CX/8= sin(a), CX/15= sin(b), and a+ b= 70. Is that enough?
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