A) \[90{}^\circ \]
B) \[{{60}^{\text{o}}}\]
C) \[45{}^\circ \]
D) \[15{}^\circ \]
E) \[30{}^\circ \]
Correct Answer: E
Solution :
Let\[{{m}_{1}}=\]slope of\[\sqrt{3}x-y-2=0\]i.e., \[{{m}_{1}}=\sqrt{3}\] and\[{{m}_{2}}=\]slope of\[x-\sqrt{3}y+1=0\]i.e., \[{{m}_{2}}=\frac{1}{\sqrt{3}}\] \[\therefore \] \[\tan \theta =\left| \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|=\left| \frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+1} \right|=\left| \frac{3-1}{2.\sqrt{3}} \right|\] \[=\frac{1}{\sqrt{3}}\] \[\therefore \] \[\theta =30{}^\circ \]
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