What is the angle in degrees between the line | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The given lines are $y = 1x + 5$ and $y = \sqrt{3}x - 4$. Comparing these with $y = mx + c$,we get slopes $m_1 = 1$ and $m_2 = \sqrt{3}$. The angl

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(D) The given lines are $y = 1x + 5$ and $y = \sqrt{3}x - 4$. Comparing these with $y = mx + c$,we get slopes $m_1 = 1$ and $m_2 = \sqrt{3}$. The angle $	heta$ between two lines is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$. Substituting the values: $	an 	heta = \left| \frac{1 - \sqrt{3}}{1 + (1)(\sqrt{3})} ight| = \left| \frac{1 - \sqrt{3}}{1 + \sqrt{3}} ight|$. Rationalizing the expression: $	an 	heta = \left| \frac{(1 - \sqrt{3})^2}{1 - 3} ight| = \left| \frac{1 + 3 - 2\sqrt{3}}{-2} ight| = \left| \frac{4 - 2\sqrt{3}}{-2} ight| = | -2 + \sqrt{3} | = 2 - \sqrt{3}$. Since $	an(15^\circ) = 2 - \sqrt{3}$,we have $	heta = 15^\circ$.

What is the angle in degrees between the lines $y = x + 5$ and $y = \sqrt{3}x - 4$ (in $^\circ$)?

Similar Questions

The slope of a line $L$ is $2$. If $m_1$ and $m_2$ are the slopes of two lines which are inclined at an angle of $\frac{\pi}{6}$ with $L$,then $m_1 + m_2 =$

If the two lines $7x + 3y + 9 = 0$ and $y = kx + 7$ are perpendicular to each other,find the value of $k$.

The angle between the straight lines $x - y\sqrt{3} = 5$ and $\sqrt{3}x + y = 7$ is ............. $^\circ$.

Find the equation of the lines passing through the origin and making an angle of $\tan^{-1} m$ with the line $y = mx + c$.

The distance of a point $(2, 5)$ from the line $3x + y + 4 = 0$ measured along the lines $L_1$ and $L_2$ are the same. If the slope of line $L_1$ is $\frac{3}{4}$,then the slope of the line $L_2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module