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

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Homework Help Overview

The discussion revolves around finding the length of CX in a triangle with a specified area of 56.4. The problem involves advanced trigonometric concepts and the relationships between the sides and angles of the triangle.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the use of the cosine law and right triangle relationships. There is a clarification about the specific length being sought (BX) and the right angle at CXB. The original poster questions the applicability of the cosine rule given the known lengths and angles.

Discussion Status

The discussion is active, with participants providing different approaches to the problem. Some guidance has been offered regarding the relationships between the sides and angles in the context of right triangles, but no consensus has been reached on a specific method to find CX.

Contextual Notes

There is a mention of the area of the triangle and the specific lengths and angles involved, which may influence the methods discussed. The original poster indicates a limitation in the information available for applying certain trigonometric laws.

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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