The lengths of the three sides of a triangle (not necessarily a right triangle) are 3.16 meters, 8.25 meters and 10.4 meters. what is the cosine of the angle opposite the side of length 10.4 meters

For non-right triangles you must use the "Law of Cosines"  and then, the "Law of Sines" to solve this.

a=  8.25m
b= 10.4m
c= 3.16m
∠A= UNKNOWN
∠B= UNKNOWN
∠C=UNKNOWN

Law of Cosines:
c²= a²+b²-2abCos(C)
(3.16)²= (8.25)²+(10.4)²- 2(8.25)(10.4)(cos(C))
9.9856 = 68.0625 + (108.16) - (171.6)(cos(C)
9.9856 = 176.2225- 171.6 cos C
-166.2369=  - (171.6(cosC))
cosC= 0.968746503
Take the inverse cosine of that to get the measure of angle C
∠C= 15.95813246°

Now Use law of sines to find ∠B:




(take the inverse sine to get the measure of ∠B)
∠B= 60.8040992°


The angle measures approximately 60.80°.

Ithink it's b sorry if i'm wrong