Find the acute angle between the straight lines y = x and y = √3x

Explanation

y = x, Gradient (m₁) = 1
y = √3x gradient (m₂) = √3
\(tan\theta = \frac{m_2 - m_1}{1+m_2 m_1}\\
tan\theta = \frac{\sqrt{3}-1}{1+\sqrt{3}\times 1}\\
=\frac{1.73 -1 }{1+1.73}=
\frac{0.73}{2.73}=0.268\)
θ = Tan^-1 (0.268)
= 15^o

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