Question

What is the measure of the acute angle formed by the lines 3x–5y=–15 and 4x+2y=7?

Answer 1

Answer:

85.6 degrees.

Step-by-step explanation:

The given equations of lines are

$3x-5y=-15$

$4x+2y=7$

We need to find the measure of the acute angle formed by these lines.

$Slope=\dfrac{-\text{Coefficient of }x}{\text{Coefficient of }y}$

Slope of given lines are

$m_1=\dfrac{-3}{-5}=\dfrac{3}{5}$

$m_2=\dfrac{-4}{2}=-2$

Angle between two lines is

$\tan \theta = \left|\dfrac{m_1-m_2}{1+m_1m_2}\right|$

$\tan \theta = \left|\dfrac{\dfrac{3}{5}-(-2)}{1+\dfrac{3}{5}(-2)}\right|$

$\tan \theta = \left|\dfrac{\dfrac{3+10}{5}}{\dfrac{5-6}{5}}\right|$

$\tan \theta = \left|\dfrac{13}{1}\right|$

$\tan \theta = 13$

$\theta = \tan^{-1}(13)$

$\theta \approx 85.6^{\circ}<90^{\circ}$

Therefore, the acute angle between given lines is 85.6 degrees.