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

AI Thread Summary
The discussion focuses on solving for the length of CX in a triangle with an area of 56.4. The user initially misphrased their question, clarifying they seek the length BX, with angle CXB being a right angle. They cannot use the cosine rule due to having only one length (15) and one angle (90 degrees). The conversation suggests using two right triangles, CXA and CXB, and applying the sine function to relate CX to the known lengths and angles. The proposed relationships involve the angles at C and their sine ratios, indicating a potential method to find CX.
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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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?
 
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