Can someone solve these three questions please. I would really appreciate it

in problems 1 to 3, find the remaining angle(s) and side(s) of each traingle if it (they) exists.
1.α = 10°, γ = 40°, side c = 2
2. a=10, side b= 7, side c =8
3. side a =1, side b =2, γ =60°

α = 10°, γ = 40°, side c = 2

10 + 40 + x = 180 ======> x = 130°

sin(C) / c = sin(α) / α

α = sin(α) * ( sin(C) / c )

α = sin(10) * ( sin(130) / 2 )

α ≈ 0.0665

sin(C) / c = sin(γ) / γ

γ = sin(40) * ( sin(130) / 2 )

γ ≈ 0.246

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

8^2 = 10^2 + 7^2 - 2 * 10 * 7 * cos(c)

64 = 100 + 49 - 140 * cos(c)

64 - 149 = - 140 * cos(c)

64 - 149 = - 140 * cos(c)

-85 = -140 * cos(c)

(85/140) = cos(c)

cos^-1( 85/140) = c

c = 52.62°

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

( sin(C) / c ) * a = sin(A)

( sin(52.62) / 8 ) * 10 = sin(A)

sin^-1( ( sin(52.62) / 8 ) * 10 ) = A

A ≈ 83.4°

180 = 83.4 + 52.62 + B

B = 43.98°

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γ^2 = a^2 + b^2 - 2 * a * b * cos(γ)

γ^2 = 1^2 + 2^2 - 2 * 1 * 2 * cos(60)

γ^2 = 1 + 4 - 4 * (1/2)

γ^2 = 5 - 2

γ = √(3)


sin(A) / a = sin(γ) / γ

sin(A) = ( sin(γ) / γ ) * a

A = sin^-1 ( ( sin(γ) / γ ) * a )

A = sin^-1 ( ( sin(60) / √(3) ) * 1 )

A = sin^-1 ( ( √(3)/2 / √(3) ) )

A = sin^-1 ( 1/2 )

A = 30°

180 = 60 + 30 + B

B = 90°

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Please let me know about the first question......

your questions does not follow the usual convention . ie angle A is opposite side a, angle B is opposite side b and angle C is opposite side c.
or a diagram is needed.