Posted by **Jasmin** on Thursday, June 3, 2010 at 10:08pm.

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**, Thursday, June 3, 2010 at 10:40pm
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**, Wednesday, February 26, 2014 at 8:53pm
Gggyy

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