Answer:
\[ c^2 = a^2 + b^2 - 2ab \cos(C) \]
Given \( a = 16 \), \( b = 19 \), and \( c = 14 \), we can substitute these values into the formula:
\[ 14^2 = 16^2 + 19^2 - 2(16)(19) \cos(C) \]
Solving for \( \cos(C) \):
\[ 14^2 = 16^2 + 19^2 - 2(16)(19) \cos(C) \]
\[ 196 = 256 + 361 - 608 \cos(C) \]
\[ 196 = 617 - 608 \cos(C) \]
\[ -421 = -608 \cos(C) \]
\[ \cos(C) = \frac{-421}{-608} \]
\[ \cos(C) = \frac{421}{608} \]
Now, to find the angle \( C \), we take the inverse cosine (arccos) of \( \frac{421}{608} \). This will give us the measure of angle \( C \) in degrees.
\[ C = \arccos\left(\frac{421}{608}\right) \]
\[ C \approx 48.76^\circ \]
Now, to find angles \( A \) and \( B \), we can use the fact that the sum of the angles in a triangle is \( 180^\circ \).
\[ A + B + C = 180^\circ \]
\[ 16^\circ + 19^\circ + 48.76^\circ = 180^\circ \]
\[ A + B \approx 180^\circ - 48.76^\circ \]
\[ A + B \approx 131.24^\circ \]
Since we know the values of two angles in the triangle, we can find the third by subtracting the sum of the other two from \( 180^\circ \).
\[ C = 180^\circ - (A + B) \]
\[ C \approx 180^\circ - 131.24^\circ \]
\[ C \approx 48.76^\circ \]
So, the angles are approximately \( A \approx 16^\circ \), \( B \approx 19^\circ \), and \( C \approx 48.76^\circ \).
