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MissPhiladelphia
Let us assign an easier nomenclature for this. Let AB=a, BC=b and CA=c. When all sides of the triangle are given, you can apply the cosine law to find the angles. The equations are:

a^2 = b^2 + c^2 -2bccosA
11^2 = 16^2 + 14^2 - 2(16)(14)cosA
448 cosA = 331
cosA = 331/448
A = arccos 331/448
A = 42.37 degrees

Another equation is:
b^2 = a^2 + c^2 -2accosB
16^2 = 11^2 + 14^2 - 2(11)(14)cosB
308cosB =61
cos B = 61/308
B = arccos 61/308
B = 78.57 degrees.

Since the sum of all angles in a triangle is 180,
C = 180 - A - B = 180 - 42.37 - 78.57
C = 59.06 degrees

Arranging from smallest to largest:

Angle A = 42.37 degrees  <  Angle C = 59.06 degrees  <  Angle B = 78.57 degrees
JeanaShupp

Answer: ∠C<∠B<∠A

Step-by-step explanation:

Given: In triangle ABC,

AB=11,BC=16, and CA=14

Clearly 11<14<16

Therefore, AB<CA<BC

We know that there is a theorem which says that "If a side of a triangle is longer than the other side, then the angle opposite the longer side will be larger than the angle opposite the shorter side".

Therefore, we have that  ∠C<∠B<∠A

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