Advanced Trigonometry: Solving for CX in a Triangle with Area 56.4 (3sf)

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david18
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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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  • Triangle ABC with vertices labelled A at top, B at right and C at bottom left. Side AC is labelled 8, side BC is labelled 15, side AB is labelled x, and the angle at C is marked 70 degrees.
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Discussion

Did you try the cosine law?

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

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