6. Given a triangle with three interior angles in arithmetic progression, the difference between the longest and shortest sides is 4 times the height on the third side. Then the largest interior angle is larger than the smallest interior angle by $\qquad$ (express using inverse trigonometric functions).


6. $\pi-\arccos \frac{1}{8}$.
In $\triangle A B C$, suppose $\angle A \geqslant \angle B \geqslant \angle C$, and the altitude from $A C$ is $h$.
$$
\begin{array}{l}
\text { Then } \angle B=60^{\circ}, \angle A+\angle C=120^{\circ}, \\
B C-A B=4 h \\
\Rightarrow \frac{h}{\sin C}-\frac{h}{\sin A}=4 h \\
\Rightarrow \sin A-\sin C=4 \sin A \cdot \sin C \\
\Rightarrow 2 \cos \frac{A+C}{2} \cdot \sin \frac{A-C}{2} \\
\quad=2(\cos (A-C)-\cos (A+C)) \\
\Rightarrow \sin \frac{A-C}{2}=2\left(\cos (A-C)+\frac{1}{2}\right) \\
\Rightarrow \sin \frac{A-C}{2}=2\left(\frac{3}{2}-2 \sin ^{2} \frac{A-C}{2}\right) \\
\Rightarrow 4 \sin \frac{A-C}{2}+\sin \frac{A-C}{2}-3=0 \\
\Rightarrow \sin \frac{A-C}{2}=\frac{3}{4} \text { or }-1(\text { rejected }) \\
\Rightarrow \cos (A-C)=1-2 \sin ^{2} \frac{A-C}{2}=-\frac{1}{8} \\
\Rightarrow \angle A-\angle C=\pi-\arccos \frac{1}{8} .
\end{array}
$$