Question

Find the angle between the 3x+5y=7,2x-y+4=0 straight lines

Answer 1

Answer:

85.6° and 94.4°

Step-by-step explanation:

y = - $\frac{3}{5}$ x + $\frac{7}{5}$

y = 2x + 4

The angle β between the lines having slope $m_{1}$ and $m_{2}$ can be find from the formula

tanβ = ± ($m_{2}$ - $m_{1}$) / (1 + $m_{1}$$m_{2}$)

tanβ = ± 13 ===> β = 85.6° and supplement angle is 94.4°

Answer 2

Answer:

Required angle is ${tan}^{ - 1} ( - 13)$

Step-by-step explanation:

Given two lines are 3x+5y=7,2x-y+4=0

We can write

$3x + 5y = 7....(1) \\ 2x - y + 4 = 0.....(2)$

From equation (1),

$5y = 7 - 3x \\ y = \frac{7}{5} - \frac{3}{5} x$

We are comparing this equation with

$y = m_1x + c_1$

and get $m_1 = - \frac{3}{5} \: and \: c_1 = \frac{7}{5}$

Again from equation (2),

$2x - y + 4 = 0 \\ y = 2x + 4$

We are comparing this equation with

$y = m_2x + c_2$

and get $m_2 = 2 \\ c_2 = 4$

We know if a is angles between $y = m_1x + c_1 \: and \: y = m_2x + c_2$

then

$tan(a) = \frac{m_2 - m_1}{1 + m_1m_2} \\ = \frac{2 + \frac{3}{5} }{1 - 2 \times \frac{3}{5} } \\ = \frac{ \frac{13}{5} }{ - \frac{1}{5} } \\ = - 13$

So, $a = {tan}^{ - 1} ( - 13)$

Therefore Required angle is

${tan}^{ - 1} ( - 13)$

Know more about angle:

https://brainly.in/question/5315290

https://brainly.in/question/2727823

#SPJ3