Question: A
52°
500m
72°
C
B
Graph shape: triangle A-B-C with A at top-left, B at bottom-left, and C at right; side A–C is labeled 500m, angle at A is 52°, and angle at C is 72°. position_hint: top-left
ab length
1. **Problem statement:** Find the length of side $AB$ in triangle $ABC$ where angle $A = 52^\circ$, angle $C = 72^\circ$, and side $AC = 500$ m.
2. **Step 1: Find angle $B$.**
Since the sum of angles in a triangle is $180^\circ$,
$$
B = 180^\circ - A - C = 180^\circ - 52^\circ - 72^\circ = 56^\circ
$$
3. **Step 2: Use the Law of Sines.**
The Law of Sines states:
$$
\frac{AB}{\sin C} = \frac{AC}{\sin B} = \frac{BC}{\sin A}
$$
We want to find $AB$, so:
$$
AB = AC \times \frac{\sin C}{\sin B}
$$
4. **Step 3: Calculate $AB$.**
Calculate the sines:
$$
\sin 72^\circ \approx 0.9511, \quad \sin 56^\circ \approx 0.8290
$$
Then:
$$
AB = 500 \times \frac{0.9511}{0.8290} = 500 \times 1.147 = 573.5 \text{ m}
$$
5. **Final answer:**
The length of side $AB$ is approximately $573.5$ meters.