Question 96836
Please help me solve this trigo problem.
A parcel of land has sides measuring 175 ft., 234 ft., 295 ft. ,and 415 ft. 
and the angle between the sides of lengths 234 ft. and 295 ft. has measure
137.1 deg. What is the measure of the angle opposite this angle, to the 
nearest tenth of a degree?
<pre><font size = 4><b>

{{{drawing(400,187.5,-100,500,-100,200,
line(0,0,415,0), line(415,0,463.2721357,168.210585),
line(463.2721357,168.210585,229.8676437,164.8942032),
line(229.8676437,164.8942032,0,0), locate(-5,0,A),locate(420,0,B),
locate(470,180,C), locate(212,190,D),locate(205,-5,"415'"), locate(220,160,"137.1°"), 
locate(445,90,"175'"), locate(341,200,"234'"), locate(60,90,"295'")
)}}}
Draw in diagonal AC.


{{{drawing(400,187.5,-100,500,-100,200,
line(0,0,415,0), line(415,0,463.2721357,168.210585),
line(463.2721357,168.210585,229.8676437,164.8942032),
line(229.8676437,164.8942032,0,0), locate(-5,0,A),locate(420,0,B),
locate(470,180,C), locate(212,190,D),locate(205,-5,"415'"),locate(220,160,"137.1°"), 

locate(445,90,"175'"), locate(341,200,"234'"), locate(60,90,"295'"),
line(463.2721357,168.210585,0,0)
 )}}}

Find the length of diagonal AC by using the law of cosines in
triangle ACD

AC² = AD² + CD² - 2·AD·CD·cos(D)
AC² = 295² + 234² - 2·295·234·cos(137.1°)
AC² = 242915.8726
AC = 492.8649539

Now use the law of cosines in triangle ABC to find angle B, 
the desired angle.

cos(B) = {{{(AB^2+BC^2-AC^2)/(2*AB*BC)}}}

cos(B) = {{{(415^2+175^2-242915.8726)/(2*415*175)}}}

cos(B) = -0.2758407752

Use inverse cosine to find angle B

    B = 106.0121251°

To the nearest tenth of a degree

    B = 106.0°

Edwin</pre>