Given a triangle $ABC$ with side $b = 14.35$ cm, side $a = 7.82$ cm, and angle $B = 115^\circ 120" = 115^\circ 2'$, we need to solve the triangle completely, which means finding the remaining side $c$ and angles $A$ and $C$.

1. Problem Description

Given a triangle $ABC$ with side $b = 14.35$ cm, side $a = 7.82$ cm, and angle $B = 115^\circ 120" = 115^\circ 2'$, we need to solve the triangle completely, which means finding the remaining side $c$ and angles $A$ and $C$.

2. Solution Steps

First, let's convert the angle $B$ into decimal degrees:

$B = 115^\circ + \frac{2'}{60'} = 115^\circ + \frac{2}{60}^\circ = 115.0333^\circ$

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

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

$\frac{7.82}{\sin A} = \frac{14.35}{\sin 115.0333^\circ}$

$\sin A = \frac{7.82 \cdot \sin 115.0333^\circ}{14.35}$

$\sin A = \frac{7.82 \cdot 0.9062}{14.35} = \frac{7.0867}{14.35} \approx 0.4939$

$A = \arcsin(0.4939) \approx 29.59^\circ$

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

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

$C = 180^\circ - A - B = 180^\circ - 29.59^\circ - 115.0333^\circ$

$C = 180^\circ - 144.6233^\circ = 35.3767^\circ \approx 35.38^\circ$

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

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

$c = \frac{b \cdot \sin C}{\sin B} = \frac{14.35 \cdot \sin 35.3767^\circ}{\sin 115.0333^\circ}$

$c = \frac{14.35 \cdot 0.5787}{0.9062} = \frac{8.29}{0.9062} \approx 9.15$ cm

3. Final Answer

$A \approx 29.59^\circ$

$C \approx 35.38^\circ$

$c \approx 9.15$ cm