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Misc-I · Q91

Q.The angle between the line √3x − y − 2=0 and x − √3y + 1=0 is (A) 15° (B) 30° (C) 45° (D) 60°

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For 3x−y−2=0\sqrt3x-y-2=0: slope =3=\sqrt3. For x−3y+1=0x-\sqrt3y+1=0: slope =13=\dfrac{1}{\sqrt3}. $\tan\theta=\left|\dfrac{\sqrt3-1/\sqrt3}{1+\sqrt3\cdot1/\sqrt3}\right|=\left|\dfrac{2/\sqrt3}{2}\ri …

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