trigonometry?

Question

1. simplify the following.
sin (90*+X) sin (180*+X)+cos (90*+X) cos (180*-X)

2. solve the following right triangle by logarithms:
A=28* 30', b=18.3

Solve the following oblique triangles

1. a=31, b=15, c=17
2. a=23.47, B=115* 30' C=20* 29'
3. a=134.2, b=84.54, B=52* 9' 11"
4. a=627.7, b=412.2, A=66* 47'

Answer 1

i think you should do your own work

Answer 2

sin (90*+X) sin (180*+X)+cos (90*+X) cos (180*-X)
cosx(-sinx)+ (-sinx)(-cosx) = 0

18.3/c = cos28.5 degrees
c = 18.3/cos28.5 degrees
So go ahead and solve by logs.

1. Use Heros formula A = sqrt(s(s-a)(s-b)(s-c)) , where s is the semiperimeter of the triangle.

2. Use law of cosines and law of sines.

3. Same as 2

4 Same as 3
Watch out for the ambiguous case.

Answer 3

i have a masters in mathematics from harvard university and whoever gave you that problem is trying to trick you. You see a triangle can't possilby contain those degrees. The axitology of a triangle converses into what is called triangular nomasity. That is the way they are created so when you imply 29 30 then 18.3 that goes against the axitology of a triangle.