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I have given two slopes $m_1 = \frac{1}{2}$ and $m_2 = 1$

While finding the angle I made use of the formula $\tan(\theta) = \frac{m_1-m_2}{1+m_1m_2}$

answer is : $\theta = \arctan(\frac{-1}{3})$

But in book the answer is $\theta = \arctan(\frac{1}{3})$.

what would be the right answer?

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  • $\begingroup$ Note that there are two angles between the lines, an acute one and an obtuse one. Using $\tan \theta = \left|\dfrac{ m_1 - m_2}{1+m_1m_2} \right|$ gives you the acute angle. $\endgroup$ – Ragib Zaman May 21 '14 at 13:43
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    $\begingroup$ It depends on which angle you are looking at. It's a simple matter of "angle of slope 1 to slope 2" versus "angle of slope 2 to slope 1". If you look at it graphically, you probably want a positive angle. $\endgroup$ – orion May 21 '14 at 13:43
  • $\begingroup$ So in above question what would be the right answer..? as i found these slopes from two curves. $\endgroup$ – zonnie May 21 '14 at 13:46
  • $\begingroup$ Look at the two lines (tangent lines) and the point where they intersect. You want an acute angle measure -- going counterclockwise around the point of intersection, which line do you get to first before the acute angle? That slope needs to be $m_1$. If you go clockwise, you get the other angle measure, still "acute" but negative measure because clockwise, or equivalently (as far as arctan can tell) the obtuse angle counterclockwise. $\endgroup$ – user128390 May 21 '14 at 20:29
  • $\begingroup$ Its not clear yet.. Can you guide me. If the point of intersection of two curves is (1,1) and I have found the slope m1 = 1/2 and m2 = 1. Now what would be the angle? $\endgroup$ – zonnie May 22 '14 at 14:06
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$$ \frac{1-\frac12}{1+1\cdot\frac 12} = \frac 1 3, \qquad \frac{\frac12-1}{1+\frac12\cdot1} =-\frac13. $$ Either of these is the angle between those two lines. Where two lines intersect they form angles with measures adding up to $180^\circ$. If $\alpha+\beta=180^\circ$ then $\tan\alpha = - \tan\beta$.

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Correct formula with sign convention positive for $\theta$ counter-clockwise rotation from direction of radius vector $1$ to $2$ is

$$\tan(\theta) = \dfrac{m_2-m_1}{1+m_1m_2}.$$

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You might not observed the modules in the formula. They remove the negative signs. So the answer you get | -1/3 | becomes +1/3. So the answer in the book is correct.

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The solution given in the book is correct. The actual formula says, one needs to take the absolute value of (m1-m2)/(1+m1.m2), before you perform a tan inverse (arctan), in order to evaluate the final value of angle between two line segments. Magnitude OR Absolute value OR Modulus of any number is always positive (it is without regard to it's sign, negative or positive). So, it ALWAYS will be arctan of some positive number.

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