Hello there. To solve this question, we'll need to remember how to use the law of sines.
Say we have the following triangle:
We know that a = 16, c = 20 and A = 42º
Assuming that this is the right orientation for this triangle, I mean, the angle A is opposite to the side a and the angle C is opposite to the side c, we can make use of the law of sines:
We'll have:
16/sin(42º) = 20/sin(C)
Divide both sides of the equation by a factor of 4
4/sin(42º) = 20/sin(C)
Considering we have a triangle, none of the angles can be zero.
Such that sin(any angle) greater than zero.
We'll make the following inequality, to make our lifes easier to deal with the sines:
sin is an injective function, such that if you have two angles a1 and a2, with a2 > a1, then sin(a2) > sin(a1)
Which means that sin(45º) > sin(42º)
Inverting the inequality, we'll have:
1/sin(42º) > 1/sin(45º)
Multiply both sides by 4
4/sin(42º) > 4/sin(45º)
But we know that 4/sin(42º) = 5/sin(C), such that:
5/sin(C) > 4/sin(45º)
Divide both sides of the equation by a factor of 5
1/sin(C) > (4/5)/sin(45º)
Inverting the inequality once again, we'll have
sin(C) < (5/4) * sin(45º)
Knowing that sin(45º) = 1/sqrt(2) approx. 0.707, we get
0 < sin(C) < 0.883
In which 0.883 < 1, such that
0 < sin(C) < 1
Then
0 < C < 90º
C is an acute angle.
