If m<A = 45, AB = 10, and BC = 8, the greatest number of distinct triangles that can be constructed is:

1) 1
2) 2
3) 3
4) 0

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To determine the greatest number of distinct triangles that can be constructed using the given information, we need to apply the triangle inequality theorem. According to this theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

In this case, we have m<A = 45, AB = 10, and BC = 8. Let's consider the three possible combinations of these side lengths:

1) AB + BC = 10 + 8 = 18
The length of the third side must be greater than 18 - 10 = 8. However, we already have BC = 8, so this combination cannot form a triangle.

2) AB + AC = 10 + AC
Since we don't have any information about AC, we can't determine whether this combination satisfies the triangle inequality theorem. It is therefore not possible to form any triangles with this combination.

3) BC + AC = 8 + AC
Similarly, since we don't have any information about AC, we can't determine whether this combination satisfies the triangle inequality theorem. It is not possible to form any triangles with this combination either.

From the above analysis, we can see that it is not possible to form any triangles using the given side lengths. Therefore, the correct answer is 4) 0.

Angle A=45, AB=10, and BC=8.

This is an SSA configuration which could result in 0, 1 or two distinct triangles, depending on the length of the "dangling" leg BC, i.e. the side which does not touch the given angle A.

To determine which case applies, we construct a triangle where the side BC' cuts AC' at C' and make a right angle with AC'. Calculate the length BC'.

In this case angle C=90°, therefore angle B=45°. From an isosceles triangle with AB as the hypotenuse, BC'=5√2=7.071

Since BC=8 is greater than BC', there are two distinct triangles where the two possible positions of C are on each side of C'.

If