Question

In the following exercise, two sides and an angle are given. First determine whether the information results in no triangle, one triangle, or two triangles. Solve the resulting triangle.

a=229,b=244, and A = 37.8°

Answer 1

By the Law of Sines,

`(\sin B)/(b)=(\sin A)/(a) \\ \\ \sin B=(b \sin A)/(a) \\ \\ =(244\sin 37.8^(\circ))/(229) \\ \\ B \approx 37.42^(\circ), 142.58^(\circ)`

Of these, only 37.42° is a possible value of B (sum of angles of a triangle is 180°).

So, we have that:

`a=229, b=244, A=37.8^(\circ), B=\arcsin \left((244\sin 37.8^(\circ))/(229) \right)`

Angles of a triangle add to 180°, so:

`C=180^(\circ)-A-B=142.2^(\circ)-\arcsin \left((244\sin 37.8^(\circ))/(229) \right)</p><p>`

Using the Law of Sines again,

`(c)/(\sin C)=(a)/(\sin A) \\ \\ c=(a \sin C)/(\sin A) \\ \\ =(244\sin \left(142^(\circ)-\arcsin \left((244 \sin 37.8^(\circ))/(229) \right) \right))/(\sin 37.8^(\circ))`

So, there is one possible triangle, where:

`a=229 \\ \\ b=244 \\ \\ c=(244\sin \left(142^(\circ)-\arcsin \left((244 \sin 37.8^(\circ))/(229) \right) \right))/(\sin 37.8^(\circ)) \\ \\ A=37.8^(\circ) \\ \\ B=\arcsin \left((244\sin 37.8^(\circ))/(229) \right) \\ \\ C=142.2^(\circ)-\arcsin \left((244\sin 37.8^(\circ))/(229) \right)`