We are given a triangle $ABC$ with side $b = 14.35$ cm, side $a = 7.82$ cm, and angle $B = 115^{\circ} 120'$ . Since $120' = 2^{\circ}$, $B = 115^{\circ} + 2^{\circ} = 117^{\circ}$. We are asked to solve the triangle completely, which means finding all the angles and sides.

1. Problem Description

We are given a triangle $ABC$ with side $b = 14.35$ cm, side $a = 7.82$ cm, and angle $B = 115^{\circ} 120'$ . Since $120' = 2^{\circ}$, $B = 115^{\circ} + 2^{\circ} = 117^{\circ}$. We are asked to solve the triangle completely, which means finding all the angles and sides.

2. Solution Steps

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

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

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

$\sin A = \frac{7.82 \sin(117^{\circ})}{14.35}$

$\sin A = \frac{7.82 \times 0.8910}{14.35} = \frac{6.9676}{14.35} \approx 0.4856$

$A = \arcsin(0.4856) \approx 29.05^{\circ}$

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

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

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

$C = 180^{\circ} - 29.05^{\circ} - 117^{\circ}$

$C = 180^{\circ} - 146.05^{\circ} = 33.95^{\circ}$

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

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

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

$c = \frac{14.35 \sin(33.95^{\circ})}{\sin(117^{\circ})}$

$c = \frac{14.35 \times 0.5588}{0.8910} = \frac{8.0185}{0.8910} \approx 8.999 \approx 9.00$ cm

3. Final Answer

$A \approx 29.05^{\circ}$

$C \approx 33.95^{\circ}$

$c \approx 9.00$ cm