Question 1159444
Solve triangle ABC. (If an answer does not exist, enter DNE. Round your answers to one decimal place. Below, enter your answers so that 
∠A1 is smaller than ∠A2.)
b = 128,    c = 162,    ∠B = 46°
<pre>Unless of course, you feel like TORTURING yourself, then you can go by the other person's more complex, BORING, and time-consuming method.
Otherwise, this is the correct way to do this! 
To find &#8737C, use LAW of SINES, or: {{{matrix(1,7, b/sin (B), "=", c/sin (C), "=====>", 128/sin (46), "=", 162/sin (C))}}}
After solving, you'll  find &#8737C to be 65.5626863<sup>o</sup>, or about 66<sup>o</sup>
Now, as sin is > 0 in the 2nd quadrant, &#8737C can ALSO be 180<sup>o</sup> - 66<sup>o</sup>, or 114<sup>o</sup>.

If &#8737C is 66<sup>o</sup>, and with &#8737B being 46<sup>o</sup>, then &#8737A MUST be 68<sup>o</sup> (180<sup>o</sup> - 66<sup>o</sup> - 46<sup>o</sup>).
If &#8737C is 114<sup>o</sup>, and with &#8737B being 46<sup>o</sup>, then &#8737A MUST be 20<sup>o</sup> (180<sup>o</sup> - 114<sup>o</sup> - 46<sup>o</sup>).
Now, it's given that &#8737A<sub>1</sub> < &#8737A<sub>2</sub>, then it goes without saying that &#8737A MUST measure the SMALLER of the 2, or 20<sup>o</sup>. 
We now know that: {{{highlight_green(matrix(3,4, ANGLE, C, "=", 114^o, ANGLE,  B, "=", 46^o, ANGLE, A, "=", 20^o))}}}
You can now use any of the above angles as well as any corresponding side, along with the LAW of SINES, to get side a.
In other words, {{{matrix(1,7, a/sin (A), "=", b/sin (B), or, a/sin (A), "=", c/sin (C))}}}
That's IT!!