Question 24

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Jambmaths question: 

Find the angle between the straight line $y=x\,\text{and }y=\sqrt{3}x$

Jamb Maths Solution: 

$\begin{align}  & \text{The acute angle formed is given by}\tan \theta =\left| \frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}} \right| \\ & \text{Where }{{m}_{\text{1}}}\text{ and }{{m}_{\text{2}}}\text{are slopes of respective lines},\text{ } \\ & \text{while the general equation of a straight line is given by} \\ & y=mx+c \\ & y=x\{{{m}_{1}}=1\} \\ & y=\sqrt{3}x\{{{m}_{2}}=\sqrt{3}) \\ & \tan \theta =\left| \frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}} \right| \\ & \tan \theta =\left| \frac{\sqrt{3}-1}{1+\sqrt{3}} \right| \\ & \text{Note} \\ & \tan {{15}^{\circ }}=\tan (60-45)=\frac{\tan {{60}^{\circ }}-\tan {{45}^{\circ }}}{1+\tan {{60}^{\circ }}\tan {{45}^{\circ }}}=\frac{\sqrt{3}-1}{1+\sqrt{3}} \\ & \therefore \tan {{15}^{\circ }}=\frac{\sqrt{3}-1}{1+\sqrt{3}} \\ & \theta ={{\tan }^{-1}}\left( \frac{\sqrt{3}-1}{1+\sqrt{3}} \right)={{15}^{\circ }} \\\end{align}$

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