1. Let us first find the missing interior angle.
m<C+53+61=180
m<C=66 degrees
law of sines: [tex]\frac{sin(A)}{a} =\frac{sin(B)}{b} =\frac{sin(C)}{c}[/tex]
a,b, and c represents the lengths of the triangles and A,B, and C represents the measures of the interior angles opposite to the sides.
For the purposes of this problem let us use a shorthand version for the law of sines.
[tex]\frac{sin(B)}{b} =\frac{sin(C)}{c}\\\frac{sin(61degrees)}{b} =\frac{sin(66degrees)}{142}\\b=142sin(61degrees)/sin(66degrees)\\[/tex]
b=134.95 meters approx.
AC is equal to length b in this case so AC=134.95 meters approx.
2. law of cosines: a²=b²+c²-2bccos(A)
Let us rearrange this formula so that we can solve for cos(C) in terms of a, b, and c (the sides lengths of the triangle).
a²=b²+c²-2bccos(A)
a²-b²-c²=-2bccos(A)
[tex]\frac{a^{2}-b^{2}-c^{2}}{-2bc}[/tex]=cos(A)
cos(A)=[tex]\frac{a^{2}-b^{2}-c^{2}}{-2bc}[/tex]
Now, because we want to find measure of angle A...
cos(A)=[tex]\frac{13.7^{2}-12.2^{2}-22.1^{2}}{-2(12.2)(22.1)}[/tex]
cos(A)=0.83 approx.
A=33.52 degrees approx.
3. Solving with law of cosines.
law of cosines: a²=b²+c²-2bccos(A)
a²=16²+18²-2(16)(18)cos(52 degrees)
a²=256+324-576cos(52 degrees)
a²=580-576cos(52 degrees)
a²=225.38 approx.
a=15.01 units approx.
Solving with law of sines.
law of sines: [tex]\frac{sin(A)}{a} =\frac{sin(B)}{b} =\frac{sin(C)}{c}[/tex]
For the purposes of this problem let us use a shorthand version for the law of sines.
[tex]\frac{sin(71 degrees)}{18} =\frac{sin(52 degrees)}{x}\\xsin(71 degrees)=18sin(52degrees)\\x=\frac{18sin(52degrees)}{sin(71 degrees)} \\[/tex]
x=15.00 un approx.
