Problem Analysis:

We are solving for:

1. $b$: The length of the other leg of the right triangle.
2. $A$: The angle opposite side $a$ (in degrees).
3. $B$: The angle opposite side $b$ (in degrees).

Given:

• $a = 6$ (side opposite angle $A$),
• $c = 19$ (the hypotenuse),
• Triangle $\triangle ABC$ is a right triangle ($C = 90^\circ$).

Solution:

1. Solve for $b$ (using the Pythagorean theorem):

$c^2 = a^2 + b^2$ $b^2 = c^2 - a^2$ $b = \sqrt{c^2 - a^2}$ Substitute $c = 19$ and $a = 6$: $b = \sqrt{19^2 - 6^2} = \sqrt{361 - 36} = \sqrt{325} \approx 18.0278$

2. Solve for $A$ (using trigonometry, sine function):

$\sin A = \frac{a}{c}$ $A = \arcsin\left(\frac{a}{c}\right)$ Substitute $a = 6$ and $c = 19$: $A = \arcsin\left(\frac{6}{19}\right) \approx \arcsin(0.3158) \approx 18.3664^\circ$

3. Solve for $B$ (using the angle sum of a triangle):

In a triangle: $A + B + C = 180^\circ$ Since $C = 90^\circ$: $B = 90^\circ - A$ Substitute $A = 18.3664^\circ$: $B = 90^\circ - 18.3664^\circ \approx 71.6336^\circ$

Final Answers:

1. $b \approx 18.0278$
2. $A \approx 18.3664^\circ$
3. ( B \approx 71.6336^\circ