Question

In triangle MNO, MN=20, NO=15, and MO=18. What is the approximate difference between the largest and smallest angle measures?

Answer 1

A) 2 degrees is the answer

Answer 2

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!