Question: Find the length of line segment $BC$ in the diagram below:
A
$68^\circ$
$10$ km
$20$ km
B
C
Triangle $ABC$, with $A$ at the top center, $B$ at the bottom left, and $C$ at the bottom right; side $AB$ is labeled $10$ km, side $AC$ is labeled $20$ km, and angle $A$ is $68^\circ$; position_hint: centerUser: help
1. **State the problem:** We need to find the length of side $BC$ in triangle $ABC$ where $AB=10$ km, $AC=20$ km, and angle $A=68^\circ$.
2. **Formula used:** We use the Law of Cosines which states:
$$BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(A)$$
3. **Substitute known values:**
$$BC^2 = 10^2 + 20^2 - 2 \times 10 \times 20 \times \cos(68^\circ)$$
4. **Calculate each term:**
$$BC^2 = 100 + 400 - 400 \times \cos(68^\circ)$$
5. **Evaluate $\cos(68^\circ)$:**
$$\cos(68^\circ) \approx 0.3746$$
6. **Continue calculation:**
$$BC^2 = 500 - 400 \times 0.3746 = 500 - 149.84 = 350.16$$
7. **Find $BC$ by taking the square root:**
$$BC = \sqrt{350.16} \approx 18.71$$
**Final answer:** The length of segment $BC$ is approximately $18.71$ km.