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

1. Problem Description

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

2. Solution Steps

We can use the Law of Sines to find the angle $A$:

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

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

$\sin A = \frac{7.82 \cdot \sin 115^\circ}{14.35}$

$\sin A = \frac{7.82 \cdot 0.9063}{14.35}$

$\sin A = \frac{7.0883}{14.35}$

$\sin A = 0.4939$

$A = \arcsin(0.4939)$

$A \approx 29.60^\circ$

Now we can find angle $C$ using the fact that the sum of angles in a triangle is $180^\circ$:

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

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

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

$C = 35.40^\circ$

Now we can use the Law of Sines again to find side $c$:

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

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

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

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

$c = \frac{8.3134}{0.9063}$

$c \approx 9.17$ cm

3. Final Answer

$A \approx 29.60^\circ$

$C \approx 35.40^\circ$

$c \approx 9.17$ cm