Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Knowing points P1,P2,P3 and distance d and the angles shown in the figure, angle between a and b not necessary 90º

What's the size of K?

share|cite|improve this question
possible duplicate of What's the size of K in this figure? – Ross Millikan Mar 30 '12 at 23:49
Is not the same problem, please read it again. – Gabriel Mar 30 '12 at 23:51
I agree it is not quite the same problem. Sorry. If you are the same user, please ask that the accounts be merged. – Ross Millikan Mar 30 '12 at 23:59

It may be convenient to represent this in the complex plane, with $P2 = 0$. Let the three vertices on the top of the figure be $P4$, $P5$, $P6$ from left to right. Now $P4 = P1 (1 - i d/a)$, $P6 = P3 (1 + i d/b)$. $P5 = P4 - s P1 = P6 - t P3$ where $s$ and $t$ are real. Solve $s P1 - t P3 = P4 - P6$ and $s \overline{P1} - t \overline{P3} = \overline{P4-P6}$, obtaining $s$ and $t$ and thus $P5$. Then $k = |P5 - P2|$.

share|cite|improve this answer

Knowing $P_1,P_2,P_3$, you can use the law of cosines to work out the angle at $P_2$. Drop a perpendicular from $P_2$ to a point $Q$ on the line segment parallel to the segment of length $a$, and another to the point $R$ on the line segment parallel to the segment of length $b$. This makes two congruent little right triangles, so you can work out the angle of either one of them at $P_2$. Now for these little right triangles you have an angle and a side, $d$, so you can get the hypotenuse, $k$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.