
Dayv O. answered 05/06/22
Attentive Reliable Knowledgeable Math Tutor
always use law of cosine for SSA problems first to determine whether there is 0,1, or 2 triangles valid.
here 62=c2+82-2c*8*cos(35) si quadradic in c
corrections involve more careful attention to decimals in calculator
two triangles, c1=2.69
c2=10.41
now use law of sin to compute angle C (angle B=180-A-C)
the following is the result.
traingle1, a=6, b=8, c=2.69 Angle A=35 degrees, B=130.1 degrees, C=14.9 degrees
triangle 2, a=6, b=8, c=10.41 Angle A= 35 degrees, B=60 degrees, C= 85 degrees
check work, (sin(35))/6=.096,,,(sin(130.1))/8=.096,,,(sin(14.9))/2.69=.096
check work, (sin(35))/6=.096,,,(sin(60))/8 not close to .096 so there is something wrong with triangle 2
the problem is while angle C was computed to be 85 degrees, from law of sin, it is by nature of sin that sin (180-85)=sin(95)=sin(85)
let's try triangle 2, a=6, b=8, c=10.41 Angle A=35 degrees, B=50 degrees, C=95 degrees
check work, (sin(35))/6=.096,,,(sin(50))/8=.096,,,(sin(95))/10.41=.096
final answer
traingle1, a=6, b=8, c=2.69 Angle A=35 degrees, B=130.1 degrees, C=14.9 degrees
triangle 2, a=6, b=8, c=10.41 Angle A= 35 degrees, B=50 degrees, C= 95 degrees