MHT CET · Maths · Straight Lines
The acute angle between the lines given by \(y-\sqrt{3} x+1=0\) and
\(\sqrt{3} y-x+7=0\) is
- A \(75^{\circ}\)
- B \(60^{\circ}\)
- C \(45^{\circ}\)
- D \(30^{\circ}\)
Answer & Solution
Correct Answer
(D) \(30^{\circ}\)
Step-by-step Solution
Detailed explanation
Slope of given lines are \(m_{1}=\sqrt{3}\), and \(m_{2}=\frac{1}{\sqrt{3}}\)
\(
\begin{array}{l}
\therefore \tan \theta=\left|\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\sqrt{3} \times \frac{1}{\sqrt{3}}}\right|=\left|\frac{2}{\sqrt{3}}\right| \\
\tan \theta=\frac{1}{\sqrt{3}} \Rightarrow \theta=30^{\circ}
\end{array}
\)
\(
\begin{array}{l}
\therefore \tan \theta=\left|\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\sqrt{3} \times \frac{1}{\sqrt{3}}}\right|=\left|\frac{2}{\sqrt{3}}\right| \\
\tan \theta=\frac{1}{\sqrt{3}} \Rightarrow \theta=30^{\circ}
\end{array}
\)
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