Posted by Jasmin on .
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

11th grade 
MathMate,
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 
11th grade 
Foggy,
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