We are given a triangle $ABC$ with angle $B = 25^{\circ} 36'$ and angle $C = 124^{\circ} 24'$, and side $c = 39.2$ m. We are asked to solve the triangle completely, meaning we need to find the remaining angle $A$ and sides $a$ and $b$.

1. Problem Description

We are given a triangle $ABC$ with angle $B = 25^{\circ} 36'$ and angle $C = 124^{\circ} 24'$, and side $c = 39.2$ m. We are asked to solve the triangle completely, meaning we need to find the remaining angle $A$ and sides $a$ and $b$.

2. Solution Steps

First, we need to find angle $A$. Since the sum of angles in a triangle is $180^{\circ}$, we have:

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

We are given $B = 25^{\circ} 36'$ and $C = 124^{\circ} 24'$. Substituting these values into the equation, we get:

$A + 25^{\circ} 36' + 124^{\circ} 24' = 180^{\circ}$

$A + 149^{\circ} 60' = 180^{\circ}$

Since $60' = 1^{\circ}$, we have:

$A + 149^{\circ} + 1^{\circ} = 180^{\circ}$

$A + 150^{\circ} = 180^{\circ}$

$A = 180^{\circ} - 150^{\circ}$

$A = 30^{\circ}$

Now we use the Law of Sines to find the lengths of sides $a$ and $b$. The Law of Sines states:

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

We are given $c = 39.2$ m, $A = 30^{\circ}$, $B = 25^{\circ} 36'$ and $C = 124^{\circ} 24'$.

First, we find side $a$:

$\frac{a}{\sin A} = \frac{c}{\sin C}$

$\frac{a}{\sin 30^{\circ}} = \frac{39.2}{\sin 124^{\circ} 24'}$

$a = \frac{39.2 \cdot \sin 30^{\circ}}{\sin 124^{\circ} 24'}$

$a = \frac{39.2 \cdot 0.5}{\sin 124^{\circ} 24'}$

$a = \frac{19.6}{\sin 124^{\circ} 24'} \approx \frac{19.6}{0.825} \approx 23.757$ m

Next, we find side $b$:

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

$\frac{b}{\sin 25^{\circ} 36'} = \frac{39.2}{\sin 124^{\circ} 24'}$

$b = \frac{39.2 \cdot \sin 25^{\circ} 36'}{\sin 124^{\circ} 24'}$

$b = \frac{39.2 \cdot 0.4318}{\sin 124^{\circ} 24'}$

$b = \frac{16.92656}{0.825} \approx 20.517$ m

3. Final Answer

$A = 30^{\circ}$

$a \approx 23.757$ m

$b \approx 20.517$ m