Slope of the two lines, $$m_1=1$$ and $$m_2=\sqrt{3}$$
Let angle between them = $$\theta$$
Then, $$tan(\theta)=|\frac{m_2-m_1}{1+m_1m_2}|$$
=> $$tan(\theta) = \frac{\sqrt{3}-1}{1+\sqrt{3}}$$
=> $$tan(\theta)=\frac{tan(60)-tan(45)}{1+tan(60)tan(45)}$$
=> $$tan(\theta)=tan(60-45)$$ $$\because [tan(A-B)=\frac{tanA-tanB}{1+tanAtanB}]$$
=> $$\theta = 15$$°
=> Ans - (A)
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