How many distinct triangles can be formed for which m∠X = 51°, x = 5, and y = 2? zero one two

From the law of sines, we have:

[tex] \displaystyle{ \frac{\sin \angle X}{x}= \frac{\sin \angle Y}{y}[/tex], 

where x and y are the sides opposite to angles X and Y, respectively.


Substituting the known values, we have:



[tex] \displaystyle{ \frac{51^{\circ}}{5}= \frac{\sin \angle Y}{2}[/tex], thus

                           [tex]\displaystyle{ \sin \angle Y=\frac{\sin 51^{\circ}}{5}\cdot2\approx \frac{0.777}{5}\cdot2=0.31[/tex].


Using a calculator, we can find that arcsin(0.31)=18 degrees, approximately.

We know that sine of (180-18)=162 degrees is also 0.31. But 162 and 51 degrees add up to more than 180 degrees.

Thus, there is only one triangle that can be formed under these conditions.