trigonometry help?

Question

Solve the following oblique triangles:


12. a = 627.7°, b = 412.2, A =66° 47’

Answer 1

I'm going to assume, unlike bkc99xx, that A is the angle OPPOSITE side a, not the angle between a and b.

Since you have two sides and an angle that is not between the two sides, I believe you have to use law of sines first, while being careful.

sin B/b = sin A/a

sin B/412.2 = sin(66° 47’)/627.7
sin B = 0.001464(412.2)
B = arcsin(0.6035)
B = 37.12° = 37° 7'

But remember that with law of sines you have to be really careful... because arcsin often has TWO valid values between 0° and 180°. So B MIGHT also be:

B = 180 - 37.12° = 142.88° = 142° 53'

But then you can throw that out, because if B were that large, C would be 180° - 66° 47' - 142° 53', which is negative. So no need to consider this possibility.

If B = 37° 7', then C = 180° - 66° 47' - 37° 7' = 76° 6' = 76.01°

sin A/a = sin C/c
c = a sin C/sin A = 627.7 sin(76° 6')/sin(66° 47') = 663

So the answers I get are:

a = 627.7°, A = 66° 47', b = 412.2, B = 37° 7', c = 663, C = 76° 6'.

Answer 2

Maybe I'm not familiar with the terminology here.

Do you mean a= 627.7 units, b= 412.2 units and A is the angle between a and b and is equal to 66 degrees 47'?

If so, using the law of cosine, then

c= (a^2 + B^2 - 2*a*b*cos(66.7833))^0.5

c= 599.9 units