Question 141297
Solve the triangle using Law of sines or cosines.
A=42(angle A)
b=120 (side b)
c=160 (side c)

<pre><font size = 4><b>
Keep all decimals until the end, then round.

{{{a^2=b^2+c^2-2*b*c*cosA}}}
{{{a^2=120^2+160^2-2*120*160*cos42}}}°

Make sure your calculator is in DEGREE mode
and not RADIAN mode.

Punch that in your calculator and you get:

{{{a^2=11463.2387}}}

Then taking square roots of both sides

{{{a = 107.0665153}}}

Now we use the law of sines:

{{{a/sinA=b/sinB=c/sinC}}}

We need only the first two parts:

{{{a/sinA=b/sinB}}}

cross-multiply:

{{{a*sinB = b*sinA}}}

Divide both sides by a

{{{sinB = (b*sinA)/a}}}

Substitute:

{{{sinB = (120*sin42)/107.0665153}}}

Calculate

{{{sinB = 0.7499606439}}}

No use the inverse sine key 
on your calculator

{{{B = 48.58696886}}}°

Now we find angle C from

{{{A + B + C = 180}}}°

{{{C = 180 - A - B}}}

{{{C = 180 - 42 - 48.58696886}}}

{{{C = 180 - 42 - 48.58696886}}}

{{{C = 89.41303114}}}°
  
So the three missing parts are

{{{a = 107.0665153}}}
{{{B = 48.58696886}}}°
{{{C = 89.41303114}}}°

But using the rounding rules, the angles
should be rounded to the nearest degree just 
like the given angle 42°, and the side should
be rounded to the nearest two significant 
digits.  So using the rounding rules:

{{{a = 110}}}  (side a)
{{{B = 49}}}°  (Angle B) 
{{{C = 89}}}°  (Angle C)

Edwin</pre>