missing angles and sides?

Question

I am given: alpha = 45, a=3, c=4

I have to find beta, gamma, b

please help and explain this step by step, I want to get the idea of how to do these. thanks

Answer 1

A = 45, a=3, c=4 .. . . this is an ambiguous case
by sine law
sin C = 4 sin 45 /3 = 0.94281
C = 70.529° . . . . and . .. C = (180 - C) = 109.471°
B = 180 - C - A = 64.471° . . . .and . . . B = 25.529°

by sine law again
b = sin 64.471 (3 / sin 45 ) = 3.828
the other case
b = sin 25.529 (3 / sin 45 ) = 0.9494

Answer 2

We have to assume:
It is a plane triangle.
Angle alpha is opposite side a
Beta opposite b
Gamma opposite c.

Sine Rule:

a/Sin(Alpha) = b/Sin(Beta) = c/Sin(Gamma)

we have Alpha, a and c, we can find Gamma:

Sin(Gamma) = c*Sin(Alpha)/a = 0.942809...

Gamma = 70.5288 or 109.4712
(Sine is symmetrical about 90)

Next we use the inner sum of angles = 180.

If Gamma = 70.5288, then
Beta = 180 - Alpha - Gamma =
180 - 45 - 70.5288 = 64.4712

Back to the sine rule
b = a* Sin(Beta) / Sin(Alpha)
b = 3*SIn(64.4712)/Sin(45) = 3.8284

If Gamma = 109.4712, then
Beta = 180 - Alpha - Gamma =
180 - 45 - 109.4712 = 25.5288

Back to the sine rule
b = a* Sin(Beta) / Sin(Alpha)
b = 3*SIn(25.5288)/Sin(45) = 1.8284

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Answer 3

use sine and cosine rule

sin 45/3 = sin gamma/4
sin(gamma)=0.94
gamma = 70.5

beta = 180 - 45 - 70.5 = 64.47

b^2 = a^2 + c^2 - 2ac cos(beta)
b^2 = 3^2 + 4^2 - 2*3*4*cos(64.47)
= 14.65
b = 3.83