The portion of the line 4x+5y=20 in the first quadrant is trisected by the lines L1 and L2 passing through the origin. The tangent of an angle between the lines L1 and L2 is:
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Official solution
The line 4x+5y=20 meets the axes at (5,0) and (0,4). The trisection points of this segment are
\[T_1 = \left(\frac{2\cdot5+1\cdot0}3,\frac{2\cdot0+1\cdot4}3\right)=\left(\frac{10}3,\frac43\right), \quad T_2=\left(\frac{5+2\cdot0}3,\frac{0+2\cdot4}3\right)=\left(\frac53,\frac83\right)\]
Slopes of L1=OT1 and L2=OT2:
\[m_1 = \frac{4/3}{10/3}=\frac25, \qquad m_2 = \frac{8/3}{5/3}=\frac85\]
\[\tan\theta = \left|\frac{m_1-m_2}{1+m_1m_2}\right| = \left|\frac{\frac25-\frac85}{1+\frac{16}{25}}\right| = \frac{6/5}{41/25} = \frac{30}{41}\]