Question 1101251: How many triangles XYZ can be formed with angle x =30,x=3 and y=8
Found 2 solutions by ankor@dixie-net.com, greenestamps:
Answer by ankor@dixie-net.com(22740)   (Show Source):
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
The other tutor didn't read the problem correctly; the given numbers are not the lengths of the three sides of the triangle.
(By chance, his answer is right; but not for the right reason.)
This problem is one where we need to determine whether there are 0, 1, or 2 triangles that can be formed, given one angle and two sides, one of which is opposite the given angle.
I always draw the picture for this type of problem the same way: the side with unknown length is horizontal, with the given angle at the left, opening upward.
So with the given information being angle X = 30 degrees, side y=8 and side x=3, I have XY horizontal, with X at the left, and segment XZ pointing upward making angle YXZ 30 degrees.
Segment XZ is y=8; and I'm wanting segment ZY to be x=3.
But if triangle YXZ were a right triangle, with angle X being 30 degrees and side XZ having length 8, segment YZ would have length 4.
But the length of segment YZ is only 3; it is not long enough to reach side XY.
So no triangles can be formed with the given measurements.
Formally, the shortest distance from point Z to line XY is the length of the segment ZY that is perpendicular to XY, making XYZ a right triangle.
If XYZ is a right triangle, then ZY/XZ is the sine of 30 degrees, which is 1/2.
That means there would be exactly one triangle that could be formed if YZ were exactly 4 (half the length of XZ); if YZ is shorter than 4, then no triangles can be formed (YZ is not long enough to reach XY); and if YZ is longer than 4, then two different triangles can be formed, because YZ could reach XZ in two different places.
Note that it could be possible for a similar problem to say that YZ was longer than XZ; but if you draw the picture of that case, it really doesn't make any sense.
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