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If $\alpha$ and $\beta$ are the angles made by the normals drawn from the origin to the lines $x+y+\sqrt{2}=0$ and $x-\sqrt{3} y-2=0$ with the positive direction of the $X$-axis respectively measured in anti-clockwise direction, the $\alpha+\beta=$
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Verified Answer
The correct answer is:
$\frac{35 \pi}{12}$
Normal drawn from origin to lines $x+y+\sqrt{2}=0$ and $x-\sqrt{3} y-2=0$
Equation as normal $x-y=0$ and $\sqrt{3} x+y=0$
$$
\alpha=1, \beta=-\frac{1}{\sqrt{3}}
$$
So, in anticlockisise direction
$$
\tan \alpha=-1 \text { and } \tan \beta=\frac{1}{\sqrt{3}}
$$
So, $\quad 2 \pi-\frac{\pi}{4}=\alpha$ and $\beta=\pi+\frac{\pi}{6}$
$$
\alpha=\frac{7 \pi}{4} \text { and } \beta=\frac{7 \pi}{6}
$$
So, $\quad \alpha+\beta=\frac{7 \pi}{4}+\frac{7 \pi}{6}$
$$
\Rightarrow \quad \frac{21 \pi+14 \pi}{12}=\frac{35 \pi}{12}
$$
Equation as normal $x-y=0$ and $\sqrt{3} x+y=0$
$$
\alpha=1, \beta=-\frac{1}{\sqrt{3}}
$$
So, in anticlockisise direction
$$
\tan \alpha=-1 \text { and } \tan \beta=\frac{1}{\sqrt{3}}
$$
So, $\quad 2 \pi-\frac{\pi}{4}=\alpha$ and $\beta=\pi+\frac{\pi}{6}$
$$
\alpha=\frac{7 \pi}{4} \text { and } \beta=\frac{7 \pi}{6}
$$
So, $\quad \alpha+\beta=\frac{7 \pi}{4}+\frac{7 \pi}{6}$
$$
\Rightarrow \quad \frac{21 \pi+14 \pi}{12}=\frac{35 \pi}{12}
$$
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