Answer:
d) 28 degrees
Step-by-step explanation:
Be,
a = MN = 20
b = NO = 15
c = MO = 18
A = angle opposite side "a"
B = angle opposite side "b"
C = angle opposite side "c"
By the law of cosines, we know that,
\(a^{2}=b^{2}+c^{2}-2bc*CosA\)
\(b^{2}=a^{2}+c^{2}-2ac*CosB\)
\(c^{2}=a^{2}+b^{2}-2ab*CosC\)
Isolating "A" from the first equation we can clamp the angle opposite to the side "a", as follows
\(a^{2}=b^{2}+c^{2}-2bc*CosA\)
\(a^{2}-b^{2}-c^{2}=-2bc*CosA\)
\(CosA=(a^{2}-b^{2}-c^{2}) / (-2bc)\)
\(A=Cos^{-1}(a^{2}-b^{2}-c^{2}) / (-2bc)\)
Replace the values and calculate the value of angle "A", like this
\(A=Cos^{-1}(20^{2}-15^{2}-18^{2}) / (-2*15*18)\)
\(A=Cos^{-1}(400-225-324) / (-540)\)
\(A=Cos^{-1}(-149) / (-540)\)
\(A=Cos^{-1}(0.2759259)\)
A = 73.98 ~ 74 degrees
Now calculate the value of angle B in a similar way,
\(b^{2}=a^{2}+c^{2}-2ac*CosB\)
\(b^{2}-a^{2}-c^{2}=-2ac*CosB\)
\(CosB=(b^{2}-a^{2}-c^{2}) / (-2ac)\)
\(B=Cos^{-1}(b^{2}-a^{2}-c^{2}) / (-2ac)\)
Replace the values and calculate the value of angle "B", like this
\(B=Cos^{-1}(15^{2}-20^{2}-18^{2}) / (-2*20*18)\)
\(B=Cos^{-1}(225-400-324) / (-720)\)
\(B=Cos^{-1}(-499) / (-720)\)
\(B=Cos^{-1}(0.6930555)\)
B = 46.13 ~ 46 degrees
The sum of the angles of a triangle is 180 degrees, that is,
A + B + C = 180 degrees
Isolating C,
C = 180 - A - B
C = 180 - 74 - 46
C = 60
Being A = 74, B = 46, C = 60, then the approximate difference between the major and minor angle measures is,
Difference = A - C
Difference = 74 - 46
Difference = 28 degrees
Hope this helps!
