# Intersection point and angle between the extended hypotenuses of two right-angled triangles in the plane

The end points of two line segments (the hypotenuses of red and blue right-angled triangles below) are given below with their coordinates marked separately on the $x$- and $y$-axes. These points are

• red triangle: $(0,1.13684)$ and $(1,1.13682)$
• blue triangle: $(0,1.1357)$ and $(1,1.1366)$.

Can you help to find the point of intersection $(X,Y)$ if these line segments were extended, and the acute angle $a$ between them? And please also show the working? ## 1 Answer

We can easily calculate the slope, rise/run, for the two lines. The line from the upper triangle has the slope

$$m_1=\frac{1.13682-1.13684}{1-0}=-0.00002$$

The line from the lower triangle has the slope

$$m_2=\frac{1.1366-1.1357}{1-0}=0.0009$$

The angle of inclination of a line is the arctangent of the slope, and one of the angles between two lines is the difference of their angles of inclination. Therefore, an angle between your two lines is $$\arctan m_2-\arctan m_1$$ $$=\arctan (0.0009)-\arctan (-0.00002)$$ $$=\arctan (0.0009)+\arctan (0.00002)$$ $$\approx 0.000919999757 \text{ radians}$$ $$\approx 0.052712103229°$$

• Thanks, this helps however, I was expecting and angle in degrees superior to 0.053 (3dp) degrees, Given the values of Y are small, I assume I can multiple the angle by 1000 to have something more significant. Thanks again for your help! – Andrew Oct 11 '15 at 13:14
• Thanks for your answer but can I ask you to verify the answers please. For the value of X I was expecting something between 1 and 2 and for the values of Y I was expecting something of the order of 1.1357 and 1.13684. My reasons are that the triangles relate to each other and are based on the same scales. – Andrew Oct 13 '15 at 7:36