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What is the angle between the two straight lines
$\mathrm{y}=(2-\sqrt{3}) \mathrm{x}+5$ and $\mathrm{y}=(2+\sqrt{3}) \mathrm{x}-7 ?$
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$\mathrm{y}=(2-\sqrt{3}) \mathrm{x}+5$ and $\mathrm{y}=(2+\sqrt{3}) \mathrm{x}-7 ?$
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Verified Answer
The correct answer is:
$60^{\circ}$
The given lines are $y=(2-\sqrt{3}) x+5$ and $y=(2+\sqrt{3}) x-7$
Therefore, slope of first line $\mathrm{m}_{1}=2-\sqrt{3}$ and slope of second line $\mathrm{m}_{2}=2+\sqrt{3}$ $\tan \theta=\left|\frac{\mathrm{m}_{2}-\mathrm{m}_{1}}{1+\mathrm{m}_{1} \mathrm{~m}_{2}}\right|=\left|\frac{2+\sqrt{3}-2+\sqrt{3}}{1+(4-3)}\right|$
$=\left|\frac{2 \sqrt{3}}{2}\right|=\sqrt{3}=\tan \frac{\pi}{3} \Rightarrow \theta=\frac{\pi}{3}=60^{\circ}$
Therefore, slope of first line $\mathrm{m}_{1}=2-\sqrt{3}$ and slope of second line $\mathrm{m}_{2}=2+\sqrt{3}$ $\tan \theta=\left|\frac{\mathrm{m}_{2}-\mathrm{m}_{1}}{1+\mathrm{m}_{1} \mathrm{~m}_{2}}\right|=\left|\frac{2+\sqrt{3}-2+\sqrt{3}}{1+(4-3)}\right|$
$=\left|\frac{2 \sqrt{3}}{2}\right|=\sqrt{3}=\tan \frac{\pi}{3} \Rightarrow \theta=\frac{\pi}{3}=60^{\circ}$
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