In the triangle ABC, the sides are c=9 b=12.The given angles are B=2x and C=x. Find all missing sides and angles.?

sin(B)/b = sin(C)/c …… (sine law)
sin(2x)/12 = sin(x)/9
2sin(x)cos(x)/12 = sin(x)/9
sin(x)cos(x)/sin(x) = 2/3

For sin(x) ≠ 0, then
cos(x) = 2/3
x = 48.19°

B = 2 * (48.19)
B = 96.38°

C = 48.19°

A = 180° - 96.38° - 48.19°
A = 35.43°

sin(A)/a = sin(C)/c …… (sine law)
sin(35.43°)/a = sin(48.19°)/9
a = 9*sin(35.43°)/sin(48.19°)
a = 7

OR:
a² = b² + c² - 2*b*c*cos(A) …… (cosine law)
a² = 12² + 9² - 2*12*9*cos(35.43)
a = 7

The missing side is a, and the missing angle is A.

Sin (2x)/12 = Sin (x) /9

= Sin (180 -3x)/A

We could use the law of sines

sin(C) / c = sin(B) / b
sin(x) / 9 = sin(2x) / 12
12 * sin(x) = 9 * sin(2x)
4 * sin(x) = 3 * sin(2x)
4 * sin(x) = 3 * 2 * sin(x) * cos(x)
2 * sin(x) = 3 * sin(x) * cos(x)

Now, sin(x) = 0 provides a possible solution set, but that would mean that x = 0 or x = 180 degrees. That doesn't seem reasonable, so we can go ahead and divide both sides by sin(x) and get rid of it

2 = 3 * cos(x)
2/3 = cos(x)
x = arccos(2/3)

C = arccos(2/3)
B = 2 * arccos(2/3)

A + B + C = 180
A + 2 * arccos(2/3) + arccos(2/3) = 180
A + 3 * arccos(2/3) = 180
A = 180 - 3 * arccos(2/3)
cos(A) = cos(180 - 3 * arccos(2/3))
cos(A) = cos(180)cos(3 * arccos(2/3)) + sin(180)sin(3 * arccos(2/3))
cos(A) = -1 * cos(3 * arccos(2/3)) + 0 * sin(3 * arccos(2/3))
cos(A) = -cos(3 * arccos(2/3))

cos(3t) =>
cos(2t + t) =>
cos(2t)cos(t) - sin(2t)sin(t) =>
cos(t)^2 * cos(t) - sin(t)^2 * cos(t) - 2 * sin(t)^2 * cos(t) =>
cos(t)^3 - 3 * sin(t)^2 * cos(t) =>
cos(t)^3 - 3 * (1 - cos(t)^2) * cos(t) =>
cos(t)^3 - 3 * cos(t) + 3 * cos(t)^3 =>
4 * cos(t)^3 - 3 * cos(t)

-cos(3 * arccos(2/3)) =>
3 * cos(arccos(2/3)) - 4 * cos(arccos(2/3))^3 =>
3 * (2/3) - 4 * (2/3)^3 =>
2 - 4 * (8/27) =>
2 - 32/27 =>
54/27 - 32/27 =>
22/27

cos(A) = 22/27
A = arccos(22/27)


A = arccos(22/27)
B = arccos(2/3)
C = 2 * arccos(2/3)

sin(A) / a = sin(B) / b = sin(C) / c

sin(arccos(22/27)) / a = sin(arccos(2/3)) / 12
12 * sin(arccos(22/27)) / sin(arccos(2/3)) = a
a = 12 * sqrt(1 - cos(arccos(22/27))^2) / sqrt(1 - cos(arccos(2/3))^2)
a = 12 * sqrt(1 - 484/729) / sqrt(1 - 4/9)
a = 12 * sqrt((729 - 484) / 729) / sqrt((9 - 4) / 9)
a = 12 * sqrt((729 - 484) * 9 / (729 * 5))
a = 12 * sqrt((729 - 500 + 16) / (81 * 5))
a = 12 * sqrt((745 - 500) / (81 * 5))
a = 12 * sqrt(245 / (81 * 5))
a = 12 * sqrt(49 / 81)
a = 12 * (7/9)
a = 4 * 7 / 3
a = 28/3

You can use your calculator to find the values for A , B , and C. arccos is just the inverse cosine function, which is usually represented with a cos-1 button. Make sure that you're in degree mode, and make sure that your answers sum to 180