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Sagot :
Certainly! Let's find the acute angle between the two given lines:
[tex]\[6x + 5y - 1 = 0\][/tex]
[tex]\[3y - 2x + 4 = 0\][/tex]
First, we need to convert these equations into the slope-intercept form ([tex]\(y = mx + b\)[/tex]), where [tex]\(m\)[/tex] represents the slope of the line.
1. Convert the first equation [tex]\(6x + 5y - 1 = 0\)[/tex]:
[tex]\[ 6x + 5y - 1 = 0 \][/tex]
Rearrange to solve for [tex]\(y\)[/tex]:
[tex]\[ 5y = -6x + 1 \][/tex]
[tex]\[ y = \left( -\frac{6}{5} \right)x + \frac{1}{5} \][/tex]
So, the slope [tex]\(m_1\)[/tex] of the first line is:
[tex]\[ m_1 = -\frac{6}{5} \][/tex]
2. Convert the second equation [tex]\(3y - 2x + 4 = 0\)[/tex]:
[tex]\[ 3y - 2x + 4 = 0 \][/tex]
Rearrange to solve for [tex]\(y\)[/tex]:
[tex]\[ 3y = 2x - 4 \][/tex]
[tex]\[ y = \left( \frac{2}{3} \right)x - \frac{4}{3} \][/tex]
So, the slope [tex]\(m_2\)[/tex] of the second line is:
[tex]\[ m_2 = \frac{2}{3} \][/tex]
3. Calculate the angle between the two lines:
To find the angle [tex]\(\theta\)[/tex] between two lines with slopes [tex]\(m_1\)[/tex] and [tex]\(m_2\)[/tex], we use the following formula for the tangent of the angle between two lines:
[tex]\[ \tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 \cdot m_2} \right| \][/tex]
Substitute the values of [tex]\(m_1\)[/tex] and [tex]\(m_2\)[/tex]:
[tex]\[ \tan(\theta) = \left| \frac{-\frac{6}{5} - \frac{2}{3}}{1 + \left( -\frac{6}{5} \cdot \frac{2}{3} \right)} \right| \][/tex]
Simplify the numerator:
[tex]\[ -\frac{6}{5} - \frac{2}{3} = -\frac{18}{15} - \frac{10}{15} = -\frac{28}{15} \][/tex]
Simplify the denominator:
[tex]\[1 + \left( -\frac{6}{5} \cdot \frac{2}{3} \right) = 1 + \left( -\frac{12}{15} \right) = 1 - \frac{4}{5} = \frac{1}{5} \][/tex]
Thus,
[tex]\[ \tan(\theta) = \left| \frac{-\frac{28}{15}}{\frac{1}{5}} \right| = \left| -\frac{28}{15} \times 5 \right| = \left| -\frac{140}{15} \right| = \left| -\frac{28}{3} \right| = \frac{28}{3} \][/tex]
So,
[tex]\[ \tan(\theta) = 9.33333333333333 \][/tex]
Now, to find the angle [tex]\(\theta\)[/tex], we take the arctan (inverse tangent) of [tex]\(9.33333333333333\)[/tex]:
[tex]\[ \theta = \arctan(9.33333333333333) \][/tex]
In radians this is:
[tex]\[ \theta \approx 1.464060654145761 \][/tex]
Finally, we convert this angle from radians to degrees:
[tex]\[ \theta \approx 83.8844964337146° \][/tex]
So, the acute angle between the two lines is approximately [tex]\(83.8844964337146^\circ\)[/tex].
[tex]\[6x + 5y - 1 = 0\][/tex]
[tex]\[3y - 2x + 4 = 0\][/tex]
First, we need to convert these equations into the slope-intercept form ([tex]\(y = mx + b\)[/tex]), where [tex]\(m\)[/tex] represents the slope of the line.
1. Convert the first equation [tex]\(6x + 5y - 1 = 0\)[/tex]:
[tex]\[ 6x + 5y - 1 = 0 \][/tex]
Rearrange to solve for [tex]\(y\)[/tex]:
[tex]\[ 5y = -6x + 1 \][/tex]
[tex]\[ y = \left( -\frac{6}{5} \right)x + \frac{1}{5} \][/tex]
So, the slope [tex]\(m_1\)[/tex] of the first line is:
[tex]\[ m_1 = -\frac{6}{5} \][/tex]
2. Convert the second equation [tex]\(3y - 2x + 4 = 0\)[/tex]:
[tex]\[ 3y - 2x + 4 = 0 \][/tex]
Rearrange to solve for [tex]\(y\)[/tex]:
[tex]\[ 3y = 2x - 4 \][/tex]
[tex]\[ y = \left( \frac{2}{3} \right)x - \frac{4}{3} \][/tex]
So, the slope [tex]\(m_2\)[/tex] of the second line is:
[tex]\[ m_2 = \frac{2}{3} \][/tex]
3. Calculate the angle between the two lines:
To find the angle [tex]\(\theta\)[/tex] between two lines with slopes [tex]\(m_1\)[/tex] and [tex]\(m_2\)[/tex], we use the following formula for the tangent of the angle between two lines:
[tex]\[ \tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 \cdot m_2} \right| \][/tex]
Substitute the values of [tex]\(m_1\)[/tex] and [tex]\(m_2\)[/tex]:
[tex]\[ \tan(\theta) = \left| \frac{-\frac{6}{5} - \frac{2}{3}}{1 + \left( -\frac{6}{5} \cdot \frac{2}{3} \right)} \right| \][/tex]
Simplify the numerator:
[tex]\[ -\frac{6}{5} - \frac{2}{3} = -\frac{18}{15} - \frac{10}{15} = -\frac{28}{15} \][/tex]
Simplify the denominator:
[tex]\[1 + \left( -\frac{6}{5} \cdot \frac{2}{3} \right) = 1 + \left( -\frac{12}{15} \right) = 1 - \frac{4}{5} = \frac{1}{5} \][/tex]
Thus,
[tex]\[ \tan(\theta) = \left| \frac{-\frac{28}{15}}{\frac{1}{5}} \right| = \left| -\frac{28}{15} \times 5 \right| = \left| -\frac{140}{15} \right| = \left| -\frac{28}{3} \right| = \frac{28}{3} \][/tex]
So,
[tex]\[ \tan(\theta) = 9.33333333333333 \][/tex]
Now, to find the angle [tex]\(\theta\)[/tex], we take the arctan (inverse tangent) of [tex]\(9.33333333333333\)[/tex]:
[tex]\[ \theta = \arctan(9.33333333333333) \][/tex]
In radians this is:
[tex]\[ \theta \approx 1.464060654145761 \][/tex]
Finally, we convert this angle from radians to degrees:
[tex]\[ \theta \approx 83.8844964337146° \][/tex]
So, the acute angle between the two lines is approximately [tex]\(83.8844964337146^\circ\)[/tex].
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