Step 1: Recall the slope rules for a pair of lines.
For $ax^2+2hxy+by^2=0$, if the slopes of the two lines are $m_1$ and $m_2$, then \[ m_1+m_2=-\frac{2h}{b},\qquad m_1m_2=\frac{a}{b}. \]
Step 2: Read the coefficients.
Here $2x^2-17xy+by^2=0$, so $a=2$, $2h=-17$ (that is $h=-\tfrac{17}{2}$), and the third coefficient is $b$. We are told one slope is 16 times the other, so $m_2=16m_1$.
Step 3: Use the sum of slopes.
\[ m_1+16m_1=17m_1=-\frac{2h}{b}=\frac{17}{b}\implies m_1=\frac{1}{b} \]
Step 4: Use the product of slopes to find $b$.
\[ m_1(16m_1)=16m_1^2=\frac{a}{b}=\frac{2}{b} \] Put $m_1=\frac{1}{b}$: \[ \frac{16}{b^2}=\frac{2}{b}\implies 16=2b\implies b=8 \]
Step 5: Use the angle formula.
The angle $\theta$ between the lines satisfies \[ \tan\theta=\frac{2\sqrt{h^2-ab}}{|a+b|} \] With $a=2$, $b=8$, $h=-\tfrac{17}{2}$: \[ \tan\theta=\frac{2\sqrt{\tfrac{289}{4}-16}}{10}=\frac{2\sqrt{\tfrac{225}{4}}}{10}=\frac{2\cdot\tfrac{15}{2}}{10}=\frac{15}{10}=\frac{3}{2} \]
Step 6: State the angle.
Since $\tan\theta=\frac{3}{2}$, the angle is $\tan^{-1}\!\left(\frac{3}{2}\right)$. \[ \boxed{\theta=\tan^{-1}\!\left(\tfrac{3}{2}\right)} \]