We need to construct triangle $ABC$ where $AB = 8$ cm, $\angle BAC = 40^\circ$, and $\angle ABC = 50^\circ$. We then need to measure the length of $BC$ to 1 decimal place.

1. Problem Description

We need to construct triangle $ABC$ where $AB = 8$ cm, $\angle BAC = 40^\circ$, and $\angle ABC = 50^\circ$. We then need to measure the length of $BC$ to 1 decimal place.

2. Solution Steps

First, draw the line segment $AB$ with length 8 cm.

Next, at point $A$, construct an angle of $40^\circ$.

Then, at point $B$, construct an angle of $50^\circ$.

Extend the lines from points $A$ and $B$ until they intersect. This intersection point is point $C$.

Finally, measure the length of the line segment $BC$.

Using the Law of Sines, we can find the length of $BC$.

First, we calculate the angle $\angle ACB$. Since the sum of angles in a triangle is $180^\circ$, we have:

$\angle ACB = 180^\circ - \angle BAC - \angle ABC = 180^\circ - 40^\circ - 50^\circ = 90^\circ$.

Using the Law of Sines:

$\frac{BC}{\sin(\angle BAC)} = \frac{AB}{\sin(\angle ACB)}$

$\frac{BC}{\sin(40^\circ)} = \frac{8}{\sin(90^\circ)}$

$BC = \frac{8 \cdot \sin(40^\circ)}{\sin(90^\circ)}$

Since $\sin(90^\circ) = 1$,

$BC = 8 \cdot \sin(40^\circ)$

$BC \approx 8 \cdot 0.6428 = 5.1424$ cm

Rounding to 1 decimal place, we get $BC \approx 5.1$ cm.

3. Final Answer

5.1 cm