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How do I solve the following question without using a calculator?

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    $\begingroup$ Well, you could figure out the slopes of the two lines, and then conclude something about that. $\endgroup$ – Matti P. Apr 16 '19 at 10:16
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  • Find quickly the point of intersection $S(3,4)$
  • Note that the triangle $P(1,0),Q(0,1),S(3,4)$ has a right angle at $Q$.
  • Hence, $\tan \alpha = \frac{|PQ|}{|QS|} = \frac{\sqrt{2}}{\sqrt{2}\cdot 3} = \frac{1}{3}$
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  • $\begingroup$ How can I be sure that it has a right angle at Q if the diagram is not to scale? $\endgroup$ – Abdel Rahman Shamel Apr 16 '19 at 10:33
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    $\begingroup$ The line through $(-1,0)$ and $(0,1)$ has slope 1. The line through $PQ$ has slope -1. $\endgroup$ – trancelocation Apr 16 '19 at 10:35
  • $\begingroup$ Ohhh right, that makes sense. Thank you. $\endgroup$ – Abdel Rahman Shamel Apr 16 '19 at 10:38
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    $\begingroup$ Slope $1$ means the angle at $(-1,0)$ is $45^{\circ}$. Slope $-1$ means, it is orthogonal to the line with slope $1$. $\endgroup$ – trancelocation Apr 16 '19 at 10:41
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Hint: Find the equation of two lines using the information(marked coordinates) given in the problem, the coefficient of $x$ are the slopes. Suppose the slope of the line passing the rightmost point of $X$-axis is $m_1$ and of the other line $m_2$. Then $$\tan\alpha=\Big(\frac{m_1-m_2}{1+m_1m_2}\Big)$$

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You can notice that the directing vectors of the 2 line are $\vec v_1(-1,-1,0)$ and $\vec v_2(-1,-2,0)$

From the cross product you get that $\sin (\alpha)=\frac{1}{\sqrt{10}}$

From the dot product you get that $\cos (\alpha)=\frac{3}{\sqrt{10}}$

Divide them to get $$\tan (\alpha)=\frac{1}{3}$$

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