Find Side Lengths of an Isosceles Triangle

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SUMMARY

The discussion focuses on solving for the side lengths of an isosceles triangle given the angles A=97.433°, B=41.283°, C=41.283° and a perimeter of 24.78 inches. Participants suggest using the Law of Cosines and the Law of Sines to establish relationships between the sides and angles. The equations provided include the perimeter equation p = a + b + c and various forms of the Law of Cosines. The conversation emphasizes the need to create a system of equations to find the side lengths accurately.

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Homework Statement


All three sides of an isoceles triangle are given along with its perimeter. Find the length of each side.
A=97.433 B=41.283 C=41.283
Perimeter=24.78in

Homework Equations


p = a+b+c
a2=b2+c2 - 2*b*c*cos(A)


The Attempt at a Solution


Would you somehow find the ratios of the side lengths? And then use the ratios and the perimeter to find the sides? I don't even know if this is possible. We definitely never did these kinds of triangles in geometry class. Thanks for any help :)
 
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I see you listed the law of cosines in the equations... but forgot the law of sines...
 
Law of sines: c/sin(A) = c/sin(B) = c/sin(C)

Do I have to solve this as a system of equations to solve for the sides? But where does the given perimeter come into play?
 
You have many equations:
a+ b+ c= 24.78
b= c
c^2= a^2+ b^2- 2abcos(41.283)
b^2= a^2+ c^2- 2accos(41.283)
a^2= b^2+ c^2- 2bccos(97.433)
\frac{a}{sin(97.433)}= \frac{b}{sin(41.283)}
\frac{a}{sin(97.433)}= \frac{c}{sin(41.283)}
\frac{b}{sin(41.283)}= \frac{c}{sin(41.283)}
Of course, these are not all independent.
 

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