Notes: `0.03x + 0.04y = 0.52, 0.02x - 0.05y = -0.19` Find the angle theta (in radians and degrees) between the lines

Wednesday, December 1, 2010

$\displaystyle{0.03}{x}+{0.04}{y}={0.52},{0.02}{x}-{0.05}{y}=-{0.19}$ Find the angle theta (in radians and degrees) between the lines

Given

$\displaystyle{0.02}{x}-{0.05}{y}=-{0.19}$

$\displaystyle{0.03}{x}+{0.04}{y}={0.52}$

Find the slope of line 1.

$\displaystyle{0.02}{x}-{0.05}{y}=-{0.19}$

$\displaystyle{2}{x}-{5}{y}=-{19}$

$\displaystyle-{5}{y}=-{2}{x}-{19}$

$\displaystyle{y}=\frac{{2}}{{5}}{x}+\frac{{19}}{{5}}$

The slope of line 1 is $\displaystyle{m}_{{1}}=\frac{{2}}{{5}}.$

Find the slope of line 2.

0.03x+0.04y=0.52

$\displaystyle{3}{x}+{4}{y}={52}$

$\displaystyle{4}{y}=-{3}{x}+{52}$

$\displaystyle{y}=-\frac{{3}}{{4}}{x}+{13}$

The slope of line 2 is $\displaystyle{m}_{{2}}=-\frac{{3}}{{4}}.$

Find the angle of inclination between the lines using the formula

$\displaystyle{\tan{{\left(\theta\right)}}}={\left|\frac{{{m}_{{2}}-{m}_{{1}}}}{{{1}+{m}_{{1}}{m}_{{2}}}}\right|}$

$\displaystyle{\tan{{\left(\theta\right)}}}={\left|\frac{{{\left(-\frac{{3}}{{4}}\right)}-{\left(\frac{{2}}{{5}}\right)}}}{{{1}+{\left(-\frac{{3}}{{4}}\right)}{\left(\frac{{2}}{{5}}\right)}}}\right|}$

$\displaystyle{\tan{{\left(\theta\right)}}}=\frac{{\frac{{23}}{{20}}}}{{\frac{{14}}{{20}}}}=\frac{{23}}{{14}}$

$\displaystyle\theta={\arctan{{\left(\frac{{23}}{{14}}\right)}}}={{58.7}}^{\circ}={1.0240}$ radians

The angle between the lines is 58.7 degrees or 1.0240 radians.

Notes

Wednesday, December 1, 2010

$\displaystyle{0.03}{x}+{0.04}{y}={0.52},{0.02}{x}-{0.05}{y}=-{0.19}$ Find the angle theta (in radians and degrees) between the lines

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