Let the sled have weight $W$. We apply Newton's first law by resolving forces parallel ($\hat{x}$) and perpendicular ($\hat{y}$) to the incline.

The sum of forces perpendicular to the incline is zero:

$$ \sum F_y = N - W\cos\theta = 0 \implies N = W\cos\theta $$

Since the sled is on the verge of moving up the incline, the maximum static friction force $f_s = \mu_s N$ acts down the incline. The sum of forces parallel to the incline is also zero:

$$ \sum F_x = F - W\sin\theta - f_s = 0 $$

Substituting $f_s = \mu_s N = \mu_s W\cos\theta$ and solving for $F$ gives a linear relationship:

$$ F(\mu_s) = W\sin\theta + \mu_s (W\cos\theta) $$

This equation is of the form $y = b + mx$, where $F$ is $y$ and $\mu_s$ is $x$. The y-intercept ($F$ at $\mu_s = 0$) is given as $F_1$.

$$ F_1 = W\sin\theta $$

The slope is $m = W\cos\theta$. We can find the slope from the two data points $(0, F_1)$ and $(\mu_{s2}, F_2)$:

$$ m = \frac{\Delta F}{\Delta \mu_s} = \frac{F_2 - F_1}{\mu_{s2} - 0} = W\cos\theta $$

To find $\theta$, divide the expression for $F_1$ by the expression for the slope:

$$ \frac{W\sin\theta}{W\cos\theta} = \tan\theta = \frac{F_1}{(F_2 - F_1) / \mu_{s2}} $$ $$ \tan\theta = \frac{F_1 \mu_{s2}}{F_2 - F_1} $$