We are given a triangle $ABC$ with side $b = 14.35$ cm, side $a = 7.82$ cm, and angle $B = 115^\circ 12'$. We are asked to solve the triangle completely, which means finding the remaining side $c$ and angles $A$ and $C$.

1. Problem Description

We are given a triangle $ABC$ with side $b = 14.35$ cm, side $a = 7.82$ cm, and angle $B = 115^\circ 12'$. We are asked to solve the triangle completely, which means finding the remaining side $c$ and angles $A$ and $C$.

2. Solution Steps

First, convert the angle $B$ to decimal degrees. Since there are 60 minutes in a degree, $12'$ is equal to $\frac{12}{60} = 0.2$ degrees. So, $B = 115.2^\circ$.

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

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

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

$\sin A = \frac{7.82 \sin 115.2^\circ}{14.35}$

$\sin A = \frac{7.82 \times 0.9047}{14.35} \approx \frac{7.075}{14.35} \approx 0.493$

$A = \arcsin(0.493)$

$A \approx 29.53^\circ$

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

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

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

$C = 180^\circ - 29.53^\circ - 115.2^\circ$

$C = 35.27^\circ$

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

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

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

$c = \frac{14.35 \sin 35.27^\circ}{\sin 115.2^\circ}$

$c = \frac{14.35 \times 0.5774}{0.9047}$

$c \approx \frac{8.285}{0.9047} \approx 9.158$ cm

3. Final Answer

$A \approx 29.53^\circ$

$C \approx 35.27^\circ$

$c \approx 9.16$ cm