# Solving Trigonometric Questions Without a Calculator [closed]

How do I solve the following question without using a calculator? • Well, you could figure out the slopes of the two lines, and then conclude something about that. – Matti P. Apr 16 '19 at 10:16

• 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, $$\tan \alpha = \frac{|PQ|}{|QS|} = \frac{\sqrt{2}}{\sqrt{2}\cdot 3} = \frac{1}{3}$$
• How can I be sure that it has a right angle at Q if the diagram is not to scale? – Abdel Rahman Shamel Apr 16 '19 at 10:33
• The line through $(-1,0)$ and $(0,1)$ has slope 1. The line through $PQ$ has slope -1. – trancelocation Apr 16 '19 at 10:35
• Ohhh right, that makes sense. Thank you. – Abdel Rahman Shamel Apr 16 '19 at 10:38
• Slope $1$ means the angle at $(-1,0)$ is $45^{\circ}$. Slope $-1$ means, it is orthogonal to the line with slope $1$. – trancelocation Apr 16 '19 at 10:41

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 $$\tan\alpha=\Big(\frac{m_1-m_2}{1+m_1m_2}\Big)$$

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 $$\tan (\alpha)=\frac{1}{3}$$