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
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: [tex] \frac{10.4}{Sin(B)} = \frac{3.16}{sin(15.96)} [/tex] [tex] \frac{10.4}{Sin(B)} =12.73922[/tex] [tex] 10.4 =12.73922115(sinB)[/tex] [tex]sinB= 0.816276439[/tex] (take the inverse sine to get the measure of ∠B) ∠B= 60.8040992°
