We are given a right triangle $\triangle ABC$ where:

• $\angle C = 90^\circ$,
• $\angle A = 30^\circ$,
• The length of side $AC = 8$ units.

We need to find the following:

• $\angle B$,
• The length of side $a$ (opposite to angle $A$),
• The length of side $c$ (the hypotenuse).

Step 1: Find $\angle B$

Since the sum of the angles in a triangle is always $180^\circ$, we can find $\angle B$ as follows:

$\angle B = 180^\circ - \angle A - \angle C = 180^\circ - 30^\circ - 90^\circ = 60^\circ$

So, $m\angle B = 60^\circ$.

Step 2: Use trigonometry to find side $a$

We can use the sine function for angle $A = 30^\circ$, since $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$. Here, the opposite side to $\angle A$ is side $a$, and the hypotenuse is side $c$.

$\sin(30^\circ) = \frac{a}{c}$ From trigonometric values, $\sin(30^\circ) = \frac{1}{2}$. Thus:

$\frac{1}{2} = \frac{a}{c}$ This means: $a = \frac{c}{2}$

Step 3: Use trigonometry to find side $c$

We know $AC = 8$ and angle $A = 30^\circ$. To find side $c$ (the hypotenuse), we use the cosine function for angle $A$:

$\cos(30^\circ) = \frac{AC}{c}$

We know that $\cos(30^\circ) = \frac{\sqrt{3}}{2}$, so:

$\frac{\sqrt{3}}{2} = \frac{8}{c}$

Solving for $c$:

$c = \frac{8}{\frac{\sqrt{3}}{2}} = \frac{8 \times 2}{\sqrt{3}} = \frac{16}{\sqrt{3}} = \frac{16\sqrt{3}}{3} \approx 9.24 \text{ units}$

Step 4: Find side $a$

Now, since $a = \frac{c}{2}$, we can calculate:

$a = \frac{9.24}{2} = 4.62 \text{ units}$

Final Answers:

• $m\angle B = 60^\circ$,
• $a = 4.62$ units,
• $c = 9.24$ units.

Would you like further details or clarifications?

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Here are some related questions to explore:

1. How can we verify the solution using the Pythagorean theorem?
2. What other trigonometric functions could we use to solve for the sides of the triangle?
3. How would the solution change if $\angle A$ were $45^\circ$?
4. What happens if the triangle is not a right triangle? How do we solve it then?
5. Can we use the law of sines or law of cosines for this triangle?

Tip: Always check your calculations using the Pythagorean theorem for right triangles!