\( M_{OA} = \dfrac{2}{5} \)
\( M_{OB} = \dfrac{a}{2} \)
\( = \tan \dfrac{\pi}{4} = \left| \dfrac{\dfrac{2}{5}}{1 + \dfrac{a}{2}} \right| \)
\( \Rightarrow 1 = \left| \dfrac{4 - 5a}{10 + 2a} \right| \Rightarrow a = \dfrac{-6}{7}, \dfrac{14}{3} \)
Product \( \Rightarrow -4, \text{Abs. value} = 4 \)

Consider the points \( A(5, 2) \) and \( B(2, a) \) joined by line segments from the origin \( O \). The given condition states that these segments subtend an angle \( \frac{\pi}{4} \) at the origin. Let the slopes of the lines \( OA \) and \( OB \) be \( M_{OA} \) and \( M_{OB} \) respectively.
The slope of line \( OA \) is: \[ M_{OA} = \frac{2}{5}. \]
The slope of line \( OB \) is: \[ M_{OB} = \frac{a}{2}. \]
Since the angle between the lines is \( \frac{\pi}{4} \),
we use the formula: \[ \tan \left( \frac{\pi}{4} \right) = \left| \frac{M_{OB} - M_{OA}}{1 + M_{OA} \cdot M_{OB}} \right|. \]
Given \( \tan \left( \frac{\pi}{4} \right) = 1 \),
we have: \[ 1 = \left| \frac{\frac{a}{2} - \frac{2}{5}}{1 + \frac{2}{5} \cdot \frac{a}{2}} \right|. \]
Simplifying the expression: \[ \left| \frac{5a - 4}{10 + 2a} \right| = 1. \]
Clearing the fractions: \[ \left| \frac{5a - 4}{10 + 2a} \right| = 1. \]
This gives two cases: \[ \frac{5a - 4}{10 + 2a} = 1 \quad \text{or} \quad \frac{5a - 4}{10 + 2a} = -1. \]
Case 1: \[ \frac{5a - 4}{10 + 2a} = 1. \]
Cross-multiplying: \[ 5a - 4 = 10 + 2a. \]
Rearranging terms: \[ 3a = 14 \implies a = \frac{14}{3}. \]
Case 2: \[ \frac{5a - 4}{10 + 2a} = -1. \]
Cross-multiplying: \[ 5a - 4 = -10 - 2a. \]
Rearranging terms: \[ 7a = -6 \implies a = -\frac{6}{7}. \]
The product of all possible values of \( a \) is: \[ a_1 \times a_2 = \left( \frac{14}{3} \right) \times \left( -\frac{6}{7} \right) = -4. \]
The absolute value of the product is: \[ |a_1 \times a_2| = 4. \]
Therefore: \[ 4. \]
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 