Question

What is the range of possible values for x? Diagram below. Will give medal

Answer 1

Law of COS says:

cosC=(a^2+b^2-c^2)/(2ab), where 'C' is the angle across from its side 'c'

SO:

cos88=a^2+b^2-39^2
cos80=f^2+d^2-(3x-12)^2


f=c a=d

a^2+^2=cos88+39^2
f^2+d^2=cos80+(3x-12)^2


cos88+39^2=cos80+(3x-12)^2
cos88+39^2=cos80+(9x^2-72x+144)
cos88-cos80+1377=9x^2-72x
1376.861251=9x(x-8)

9x = 1376.861251
x-8=1376.861251

x = 152.9845835
x = 1384.861251

therefore 152.984535 <= x <= 1384.861251

Answer 2

I can only choose from
a 12 < x < 17
b x greater than or equal to 5
c 4 < x < 12
d 4 < x < 17

Answer 3

First, use Trig to find the lengths of the other two sides of both triangles.

Answer 4

Sorry, I forgot that you can't use Trig functions on non-right triangles.

Answer 5

The only Trig function I can see that applies to your problem is the Law of Sines, which states that:
a/sin(A) = b/sin(B) = c/sin(C)

Answer 6

Furthermore, the sum of the lengths of any two sides of a Triangle must be greater than the length of the of the remaining side.

Spelled out, this means that:
AB + BC > AC
AB + AC > BC
BC + AC > AB

This will give you your range, when you can figure out what the lengths of the other two sides are.

Answer 7

Actually, the two triangles are unlikely to share any angles in common, either. Since the known angle is not the same in both, and the sum of the angles of all triangles totals 180 degrees, then at least one of the other two angles must differ as well.

Answer 8

So far, the Law of Sines gives us the following two equations, one for each triangle:
BC/sin(A) = 39/sin(88) = AB/sin(C)
BC/sin(D) = (3x - 12)/sin(80) = AB/sin(F)