Answer
$\bf Emily$
Work Step by Step
We know that the triangular block of Swiss cheese will not topple (rotating counterclockwise) until its center of mass passes its left corner. That occurs when the board is tilted at a critical angle of $\theta_c$, as we see in the first figure below.
We assume that the block starts sliding when the tilted angle reaches $\theta_1$.
Now we need to find the angle $\theta_1$ when it starts sliding down and at the angle $\theta_c$ when it will topple.
Intuitively, we can assume that $\theta_c\gt \theta_1$ but let's test that mathematically.
We assume that the mass in the triangular block is uniformly distributed and it is isosceles. this means that its center of mass is at its geometric center.
Hence, the line from the center of the mass to the left angle (the direction of the weight force) is dividing $\phi$ into two angles. $\measuredangle 1$ and $\measuredangle 2$.
And we can see that the block is on the verge of toppling when the sum of the two angles of $\theta_c$ and $\measuredangle 2$ is 90$^\circ$
So we have to find $\measuredangle 2$ which is given by
$$\tan\measuredangle 2=\dfrac{4}{4}$$
Since we know that the geometric center of the isosceles triangle is at one-third its height, $y_{cm}=4$ cm. Also, $x_{cm}=4$ cm.
$$\measuredangle 2=\bf 45^\circ$$
Hence, the critical angle is given by
$$\theta_c=90^\circ- \measuredangle 2=90^\circ-45^\circ$$
$$\boxed{\theta_c=\color{red}{\bf 45}^\circ}$$
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Now we need to find at what angle will the block starts to slide...
As we see in the last figure below, the block will be on the verge of sliding to the left when the weight component exceeds the static friction force between the block and the surface.
Applying Newton's second law in both directions.
$$\sum F_y=F_n-mg\cos\theta_1=ma_y=m(0)=0$$
Thus,
$$F_n=mg\cos\theta_1\tag 1$$
$$\sum F_x=mg\sin\theta_1-f_s=ma_x=m(0)=0$$
when the block is on the verge of sliding, it is not moving yet, so its acceleration is still zero.
Thus;
$$mg\sin\theta_1=f_s$$
Recalling that $f_s=\mu_sF_n$
$$mg\sin\theta_1=\mu_sF_n$$
Plugging from (1);
$$ \color{red}{\bf\not}m \color{red}{\bf\not}g\sin\theta_1=\mu_s \color{red}{\bf\not}m \color{red}{\bf\not}g\cos\theta_1$$
Hence,
$$\tan\theta_1=\mu_s=0.90$$
Therefore,
$$\boxed{\theta_1\approx \color{red}{42}^\circ}$$
Since $\theta_1\lt \theta_c$, the block will slide before topple.
So that $\bf Emily$ is right.
