Question

QF Q8.) Solve the triangle.

Answer 1

A triangle solver tells you
  C = 85°
  b ≈ 8.4
  c ≈ 10.4

From the sum of angles of a triangle, 
  C = 180° -42° -53° = 85°

From the law of sines
  b = sin(53°)/sin(42°)*7 ≈ 8.355
  c = sin(85°)/sin(42°)*7 ≈ 10.422
  

Answer 2

Comment
To start this question after finding C, you need to use the Sin Law for solving lengths (or angles)  in triangles that are not right angle triangles

That's not exactly good news. The Sin law can return no solutions, or one solution or 2 solutions. 

The programmer only gave you the opportunity of offering 1 solution.That likely means that there is only one solution or none.

Step One
Solve for C
Givens
A  = 42°
B = 53°
All triangles have 180° as the sum of their angles.

A + B + C = 180°         Substitute for A and B
42° + 53° + C = 180°   Combine the like terms on the left
95° + C = 180°            Subtract 95 from 180
C = 180° - 95°
C = 85°

Step Two
Find b
Note: You must use the Sin Law here.
Sin(A) / a = Sin(B)/b            Cross multiply.
b * Sin(A) = a* Sin(B)           Divide by Sin(A)
b = a * Sin(B) / Sin(A)

Substitute the known parts into the expression above.

b = 7 * sin(53) / Sin(42)
b = 7 * 0.7986 / 0.6691
b = 8.3548

Step Three
Find c
Comment: as soon as possible it is a good idea to switch to the cos law. It gives only one solution per a given set of numbers.

a = 7
b = 8.3548
C = 85

c^2 = a^2 + b^2 - 2*a*b * Cos(C)
c^2 = 7^2 + 8.3548^2 - 2*7*8.3548 * cos(85)  
c^2 = 49  + 69.8027 - 2*7*8.3548*0.087156
c^2 = 118.8027 - 10.1944
c^2 = 108.6083                  Take the square root of both sides.
c = sqrt(108.6083)
c = 10.4215

Answers
C = 85°
b = 8.3548
c = 10.4215 

I leave the rounding to you.

Comment
Note: we did not eliminate the possibility of no solutions. That occurs if you find the height of the triangle using (in this case) angle B. 

Sin(53) = opposite (height) / hypotenuse (which would be a in this case).
hypotenuse = 7
Sin(53) = height / hypotenuse
0.7986 = height / 7
7 * 0.7986 = height
height = 5.5902

As long as h < a (in this case it is) there is 1 solution and the 0 solutions is eliminated. Technically, a purist would insist on making this test. If a<h then a would hang like a swinging lamp from C and would never touch AB. 

We have established that there is 1 solution, but maybe there are two. No that's not possible from the givens. 2 solutions would require that 2 sides and an angle be given. The given angle cannot be between the 2 sides. There might still be only 1 solution because a > b. If it is, then a can't swing across to meet AB in 2 places.

That's why, if all possible the cos law should be used.