ANSWER:

We know that :-

\begin{gathered} \tt If \: slope \: of \:two \: lines \: are \: m_1 \: and \: m_2 then \: acute \: \: angle \: \THETA \: between \: two \: lines \: is \: given \: as : \\ \\ \sf tan\theta = \bigg | \FRAC{m_1 -m_2 }{1 +m_1 m_2 } \bigg |\end{gathered}

Ifslopeoftwolinesarem

1

andm

2

thenacuteangleθbetweentwolinesisgivenas:

tanθ=

∣

∣

∣

∣

∣

1+m

1

m

2

m

1

−m

2

∣

∣

∣

∣

∣

We also know that :-

\begin{gathered} \tt slope (m) \: \: of \: line \: \: ax + by + c = 0 \: is : \\ \\ \BOXED{ \sf m = \frac{ - a}{b} }\end{gathered}

slope(m)oflineax+by+c=0is:

m=

b

−a

\begin{gathered}\tt slope (m _1) \: \: of \: line \: \: y - \sqrt{3} x - 5 = 0 : \\ \\ \rightarrow \sf m _1 = - \: \frac{ - \sqrt{3} }{1} = \sqrt{3} \\ \\ \\ \tt slope (m _2) \: \: of \: line \: \: \sqrt{3} y - x + 6 = 0 : \\ \\ \sf \rightarrow m _2 = - \frac{ - 1}{ \sqrt{3} } = \frac{ 1}{ \sqrt{3}}\end{gathered}

slope(m

1

)ofliney−

3

x−5=0:

→m

1

=−

1

−

3

=

3

slope(m

2

)ofline

3

y−x+6=0:

→m

2

=−

3

−1

=

3

1

\begin{gathered}\sf \longrightarrow tan\theta = \bigg | \dfrac{ \sqrt{3} - \frac{1}{ \sqrt{3} } }{1 +(\sqrt{3} \times \frac{1}{ \sqrt{3}}) } \bigg | \\ \\ \sf \longrightarrow tan\theta = \bigg | \dfrac{ \frac{3 - 1}{ \sqrt{3} } }{1 +1} \bigg |\\ \\ \sf \longrightarrow tan\theta = \bigg | \dfrac{ \frac{2}{ \sqrt{3} } }{2} \bigg |\\ \\ \sf \longrightarrow tan\theta = \bigg | \frac{1}{ \sqrt{3} } \bigg |\\ \\ \sf \longrightarrow tan\theta = \frac{1}{ \sqrt{3} } \\ \\ \therefore \theta \: = 30 \degree\end{gathered}

⟶tanθ=

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1+(

3

×

3

1

)

3

−

3

1

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∣

∣

∣

∣

⟶tanθ=

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∣

∣

∣

∣

1+1

3

3−1

∣

∣

∣

∣

∣

⟶tanθ=

∣

∣

∣

∣

∣

2

3

2

∣

∣

∣

∣

∣

⟶tanθ=

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∣

∣

∣

∣

3

1

∣

∣

∣

∣

∣

⟶tanθ=

3

1

∴θ=30°

Angle between the lines y - √3x - 5 = 0 and √3y - x + 6 = 0 = 30°