# Angle between equal sides of isosceles triangle

Consider the points $$u = (1,1,−1)$$ , $$v = (a,2,−1)$$ and $$w = (1,2,−1)$$ in $$R^3$$, where $$a \in R$$. There are three possible values of $$a$$ for which $$u, v$$ and $$w$$ will form an isosceles triangle.

How do we find these values of a and hence how can we find the angle between the equal sides of the triangle?

• Once you have a value of $a$ and know which two sides are equal, the usual approach is to bisect their included angle, which reduces the calculation to trigonometry on either of the resulting right triangles. – hardmath Mar 25 at 18:29
• @hardmath How does one compute the value for a? – bigfocalchord Mar 25 at 18:51
• I'll post an Answer if no one beats me to it, but note all three points $u,v,w$ have the same $z$-coordinate, so this is about a plane triangle (and only the $x,y$ coordinates matter). – hardmath Mar 25 at 18:55
• Have you tried drawing a picture (in the plane $z=-1$)? – Andreas Blass Mar 25 at 19:28

1. $$|u-v|=|u-w|,$$ which gives $$(a-1)^2+1=1$$ or $$a=1$$, which is impossible, otherwise $$v=w$$;
2. $$|v-w|=|u-w|,$$ which gives $$(a-1)^2=1,$$ which gives two values of $$a$$.
• @bigfocalchord You can use $\cos\measuredangle(\vec{a},\vec{b})=\frac{\vec{a}\vec{b}}{|\vec{a}||\vec{b}|}.$ – Michael Rozenberg Mar 25 at 19:13