missing angle and sides?

Question

given:

gamma=40
a=3
b=4

Have to find:

Beta , alpha, c

please help me and explain this to me step by step, thanks my book sucks!!

Answer 1

Use these 3 rules

SinX=opposite/hypotenis
CosX=adjacent/hypotenis
tanX=opposite/adjacent

Substitute in these equations (there is no drawing for the traingle so i cannot solve the problem)

Answer 2

I am going to assume that gamma is the angle at C
Here's the formula: it's called "The Law of Cosines"
c^2=a^2+b^2-2abcosC
c^2=9+16-2(3)(4)cos40
c^2=25 -24(0.76604)
c^2=25-18.385
c^2=6.615
c=2.57

There are 2 other "Law of Cosines+
a^2=b^2+c^2-2bccosA
b^2=c^2+a^2-2accosB

Answer 3

b is the side opposite angle beta
a is the side opposite angle alpha
c is the side opposite angle gamma

By the law of cosines,
c^2 = a^2+b^2 - 2ab cos gamma
c^2 = 3^2 +4^2 -2(3)(4)cos(40) = 25 -18.385 = 6.6149
c = sqrt(6.6149)= 2.57

Now you can use the law of sines to find the angle alpha.
a/sin alpha = c/ sin gamma
3/sin alpha = 2.57/sin40
sin alpha = (3sin40)/2.57 = .75033
alpha = arcsin .75033 = 48.62 degrees
Thus beta = 180 -(40 +48.62) = 91.38 degrees

Answer 4

apply cosine rule

c^2 = a^2 + b^2 - 2ab cosC

c^2 = 3^2 + 4^2 - 2(3)(4) cos (40)

c^2 = 9 + 16 - 24 (0.766)

c^2 = 25 - 18.385

c^2 = 6.615

c = sqrt(6.615) = 2.572

now apply sine rule a/sin A = b/sinB = c/sinC

3/sin A = 2.572/sin 40

sinA = 3 sin (40)/2.572

sinA = 3(0.643)/2.572

sinA = 0.75

A = sin^-1(0.75)

alpha = 48.6 degrees

Beta = 180-(40+48.6) = 180 - 88.6 = 91.4 degrees