The problem asks us to draw a triangle accurately using a ruler and protractor, given the lengths of two sides (7 cm and 8 cm) and the angle between the 8 cm side and the unknown side $b$ (30 degrees). Then, we need to measure the length of side $b$ in the drawing and give the answer to one decimal place. Since it asks us to draw the diagram, I will provide an accurate answer by using the Law of Cosines.

1. Problem Description

The problem asks us to draw a triangle accurately using a ruler and protractor, given the lengths of two sides (7 cm and 8 cm) and the angle between the 8 cm side and the unknown side $b$ (30 degrees). Then, we need to measure the length of side $b$ in the drawing and give the answer to one decimal place. Since it asks us to draw the diagram, I will provide an accurate answer by using the Law of Cosines.

2. Solution Steps

The Law of Cosines states that for any triangle with sides of length $a$, $b$, and $c$, and an angle $C$ opposite side $c$:

$c^2 = a^2 + b^2 - 2ab \cos(C)$

In our case, we have sides of length 7 cm and 8 cm, and the angle between the 8 cm side and the unknown side $b$ is 30 degrees. Let $a = 8$, let $c = 7$, and let angle $C = 30$ degrees. We want to find $b$. Plugging the given values into the Law of Cosines formula, we have:

$7^2 = 8^2 + b^2 - 2(8)(b) \cos(30^\circ)$

$49 = 64 + b^2 - 16b \cos(30^\circ)$

$49 = 64 + b^2 - 16b (\frac{\sqrt{3}}{2})$

$49 = 64 + b^2 - 8\sqrt{3}b$

$0 = b^2 - 8\sqrt{3}b + 15$

Using the quadratic formula to solve for $b$:

$b = \frac{-(-8\sqrt{3}) \pm \sqrt{(-8\sqrt{3})^2 - 4(1)(15)}}{2(1)}$

$b = \frac{8\sqrt{3} \pm \sqrt{192 - 60}}{2}$

$b = \frac{8\sqrt{3} \pm \sqrt{132}}{2}$

$b = \frac{8\sqrt{3} \pm 2\sqrt{33}}{2}$

$b = 4\sqrt{3} \pm \sqrt{33}$

So, $b = 4\sqrt{3} + \sqrt{33} \approx 6.928 + 5.745 = 12.673$

or $b = 4\sqrt{3} - \sqrt{33} \approx 6.928 - 5.745 = 1.183$

However, since the angle opposite side 7 is 30 degrees, side b must be greater than side 7 because the angle opposite it is larger than 30 degrees. Therefore, $b = 4\sqrt{3} - \sqrt{33}$ must be the solution. Thus:

$b \approx 1.183$

Rounding to 1 decimal place, $b \approx 1.2$

3. Final Answer

1.2 cm