Inradius known, side unknown

The area of $\triangle ABC$ is $A = \frac{1}{2} \cdot 7 \cdot x \cdot \sin 60° = \frac{7\sqrt{3}}{4}x$ .

By the law of cosines on $\triangle ABC$ , $AB = \sqrt{x^2 + 7^2 - 2 \cdot x \cdot 7 \cdot \cos 60°} = \sqrt{x^2 - 7x + 49}$ .

The semiperimeter of $\triangle ABC$ is therefore $s = \frac{1}{2}(x + 7 + \sqrt{x^2 - 7x + 49})$ .

The relationship between the area $A$ , the inradius $r$ , and the semiperimeter $s$ is $A = rs$ . Therefore, $\frac{7\sqrt{3}}{4}x = 2 \cdot \frac{1}{2}(x + 7 + \sqrt{x^2 - 7x + 49})$ , which solves to $x = \frac{112 + 264\sqrt{3}}{83} \approx \boxed{6.859}$ .

Let $BC=a$ , $AC=b=7$ , and $AB=c$ . By cosine rule , we have $c^2 = a^2 + b^2 - 2 ab \cos 60^\circ = a^2 - 7a + 49$ . The area of $\triangle ABC$ , $[ABC] = \dfrac 12 ab \sin 60^\circ = \dfrac {7\sqrt 3}4a$ . And $[ABC] = \dfrac r2 (a+b+c) = a+7+c$ , where $r=2$ is the inradius. Therefore,

$\begin{aligned} \frac {7\sqrt 3}4a & = a+7+c \\ \left(\frac {7\sqrt 3}4 - 1\right)a - 7 & = c \\ \left(\left(\frac {7\sqrt 3}4 - 1\right)a - 7\right)^2 & = c^2 = a^2 - 7a + 49 \\ \left(\left(\frac {7\sqrt 3}4 - 1\right)^2 - 1\right)a^2 - \left(\frac {49\sqrt 3}2 - 21\right)a & = 0 & \small \blue{\text{Since } a>0} \\ \implies a & = \frac {\frac {49\sqrt 3}2 - 21}{\left(\frac {7\sqrt 3}4 - 1\right)^2 - 1} \approx \boxed{6.86} \end{aligned}$