In triangle $ABC$, we are given side $b = 14.35$ cm, side $a = 7.82$ cm, and angle $B = 115^{\circ}$. We need to solve the triangle completely, meaning we need to find the remaining angle $A$ and $C$, and the remaining side $c$.

1. Problem Description

In triangle $ABC$, we are given side $b = 14.35$ cm, side $a = 7.82$ cm, and angle $B = 115^{\circ}$. We need to solve the triangle completely, meaning we need to find the remaining angle $A$ and $C$, and the remaining side $c$.

2. Solution Steps

First, we use the Law of Sines to find angle $A$.

$\frac{a}{\sin A} = \frac{b}{\sin B}$

$\sin A = \frac{a \sin B}{b}$

$A = \arcsin(\frac{a \sin B}{b})$

$A = \arcsin(\frac{7.82 \sin(115^{\circ})}{14.35})$

$A = \arcsin(\frac{7.82 \times 0.9063}{14.35})$

$A = \arcsin(\frac{7.088}{14.35})$

$A = \arcsin(0.494)$

$A = 29.60^{\circ}$

Next, we can calculate angle $C$ using the fact that the sum of angles in a triangle is $180^{\circ}$.

$A + B + C = 180^{\circ}$

$C = 180^{\circ} - A - B$

$C = 180^{\circ} - 29.60^{\circ} - 115^{\circ}$

$C = 35.40^{\circ}$

Finally, we can find side $c$ using the Law of Sines.

$\frac{c}{\sin C} = \frac{b}{\sin B}$

$c = \frac{b \sin C}{\sin B}$

$c = \frac{14.35 \sin(35.40^{\circ})}{\sin(115^{\circ})}$

$c = \frac{14.35 \times 0.5793}{0.9063}$

$c = \frac{8.313}{0.9063}$

$c = 9.17 \text{ cm}$

3. Final Answer

$A = 29.60^{\circ}$

$C = 35.40^{\circ}$

$c = 9.17 \text{ cm}$