1. May 6, 2007

david18

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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2. May 6, 2007

HallsofIvy

Staff Emeritus
Did you try the cosine law?

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

a= 15, b= 8, and C= 70 degrees.

3. May 6, 2007

david18

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)

4. May 6, 2007

HallsofIvy

Staff Emeritus
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?