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I am sorry in advance. If I am shooting a laser in front of me. Assume 90 degrees. And then I have a second laser that I shoot from that has a point 130 degrees. The midpoint between the two is not 110 degrees, correct? I don't add them and divide by two. Because the distance between the point at 90 degrees would be shorter in distance to the point at 110 degrees, than the 110 degree point is from the point at 130 degrees. I am so sorry again, but is there a formula to use? I am not a math wizard and feel embarrassed to ask this question, but I have no idea. Thank you, thank you so much for any assistance. Even if I knew the name of the formula I could explore more.

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  • $\begingroup$ I'm not entirely sure what you're asking. Can you clarify? $\endgroup$ – Neil May 1 '16 at 4:36
  • $\begingroup$ The median of a side of a triangle is the point that divides it into two equal parts. Is this what you want? If so, you have to specify the length of the sides coming from where you are. $\endgroup$ – marty cohen May 1 '16 at 4:49
  • $\begingroup$ Thank you for answering. Suppose I am ten feet from a wall. And I pointed my laser at 90 degrees and made a mark on the wall. Then I turned my laser 40 degrees and pointed my laser and made another mark on the wall. I know that distance between the two points is a distance I can calculate (law of cosine, I believe). But suppose I wanted to move my laser again to project the exact middle between those two points from ten feet. I wouldn't move my laser to 110 degrees, would I? 110 degrees is mid angle, but that doesn't project to the midpoint. $\endgroup$ – Mike May 1 '16 at 5:06
  • $\begingroup$ Thank you for answering! To clarify, suppose I am facing a wall ten feet in front of me. And I pointed my laser straight ahead (using my protractor it reads 90 degrees) and made a mark on the wall. Then I turned my laser 40 degrees to the right (130 degrees total) and made another mark on the wall. Can I find the angle between those two (eg 113 degrees) that will result in being equidistant from the first point to the second point? If I walked 5 feet closer to the wall, would that angle change? Now that I am 5 feet closer, is the middle angle now 111 degrees instead of 113 degrees? $\endgroup$ – Mike May 1 '16 at 5:17
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Here is a diagram of your situation (note that angles in the same color are equal):

enter image description here

You are at $P$. The first laser you fire at 90 degrees meets at $A$. You know the distance $a$: for example, $a = 10\ \text{ft}$ means you are 10 feet from the wall. You also know the angle at $A$, which is 90 degrees, and the angle at $C$, which is $r$ (in your example, $r$ was 130 degrees).

There is a mathematical tool called trigonometry which allows us to find the length $AC$ from this information. We have $AC = a\cot r$, and $B$ is the midpoint of $AC$, so $AB = \frac{1}{2}a\cot r$. Now $\tan g = \frac{a}{AB} = \frac{a}{\frac{1}{2}a\cot r} = 2\tan r$ so the angle $g = \arctan(2\tan r)$.

If you don't know what most or all of these symbols mean, you should learn about trigonometry. And don't apologize, we're all here to learn!

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  • $\begingroup$ Thank you! Thank you. Now the big question. If I move closer to the wall, does angle g stay constant? If instead of being 10 feet away, I am now 5 feet away. I used my protractor and drew it up, and I think the answer is yes. What I want to do is not have to know if I am 5 or 10 feet from the wall. Just know that angle g will constant relative to angle a (90 degrees) and r (130 degrees). Will angle g be the middle between a and r if I am 5 or 10 feet away? If you know the answer quickly, I'd thank you. Otherwise I can run the calculation. Thank you again! $\endgroup$ – Mike May 1 '16 at 6:05
  • $\begingroup$ As long as the angle at $A$ and $r$ remain the same, $g$ will also remain the same. In the formula I gave, $g$ depends only on $r$ (assuming that $A$ is always 90 degrees). $\endgroup$ – shardulc May 1 '16 at 6:07
  • $\begingroup$ You are so awesome. That is what I suspected. I really, really, really appreciate this. Thank you again, shardulc! $\endgroup$ – Mike May 1 '16 at 6:11
  • $\begingroup$ @Mike You're welcome! If you like the answer, please upvote it by clicking the up-arrow, and to accept this answer, please click the tick-mark. $\endgroup$ – shardulc May 1 '16 at 6:13
  • $\begingroup$ Done and done! You really helped. Can't thank you enough. Now I can sleep! It was gnawing on me..... $\endgroup$ – Mike May 1 '16 at 6:30
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Draw a triangle, let the full angle, median angles be $ \alpha,\beta $ from central position.

$$ \tan \alpha = 2 \tan \beta. $$

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