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

User Jfenwick
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From the law of sines, we have:


\displaystyle{ (\sin \angle X)/(x)= (\sin \angle Y)/(y),

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


Substituting the known values, we have:




\displaystyle{ (51^(\circ))/(5)= (\sin \angle Y)/(2), thus


\displaystyle{ \sin \angle Y=(\sin 51^(\circ))/(5)\cdot2\approx (0.777)/(5)\cdot2=0.31.


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.


User Seminda
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