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I've been looking around, trying to find a simple way explaining why and how to calculate distance using the triangulation technique, but I'm still pretty confused, I've got some simple math notions, but I lack knowledge on using angles to solve such problems. I have a simple example, and I'd like to solve it using triangulation.

Any help is appreciated!

P.S.: I'm a layman at math, so I'm sorry in advance.

Triangle specifications:

A side: Unknown

B side: 10 meters

C(hypotenuse): Unknown

AB angle: 90°

BC angle: 70°

UPDATE:

After some searching, I found a website that clarifies:

Cos, Sin, Tan

https://www.mathsisfun.com/sine-cosine-tangent.html

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I have drawn your triangle, but the sides are named with lower case letters instead of capitals thanks to Geogebra. Side $a$ is $10 \tan (70^\circ)$ , then you can get $c$ from either $c=\sqrt {a^2+b^2}$ or $c=\frac {10}{\cos (70^\circ)}$ If you have a right angle, finding the coordinates of $B$ is pretty easy. If you don't, you need to solve a pair of simultaneous equations that represent the lengths from A to B and C to B enter image description here

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  • $\begingroup$ Thanks for the answer Ross, but I'm still a little bit confused on how to calculate it, May you go step-by-step solving the problem $\endgroup$ – Kyle Jan 12 '16 at 16:24
  • $\begingroup$ Do you know the definitions of the trig functions and understand those formulas? What coordinates are you given? Are you trying to find the coordinates of B? $\endgroup$ – Ross Millikan Jan 12 '16 at 16:38
  • $\begingroup$ The length of B is given as 10 meters in your problem. I showed how to compute the lengths of the other two sides. $\endgroup$ – Ross Millikan Jan 12 '16 at 16:48
  • $\begingroup$ I'm sorry, I meant the length of a $\endgroup$ – Kyle Jan 12 '16 at 16:49
  • $\begingroup$ In a right triangle, the tangent of an angle is defined as the opposite side divided by the adjacent side, so $\tan 70^\circ = a/b$ That is where I got $a=10 \tan 70^\circ$, but substituting in $b=10$ $\endgroup$ – Ross Millikan Jan 12 '16 at 16:52
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A good acronym that we teach in the US is "soh cah toa," which is a made-up Native American word. You can interpret it as

  • s = o/h, which means sin(angle) = "opposite side" over "hypotenuse"
  • c = a/h, which means cos(angle) = "adjacent side" over "hypotenuse"
  • t = o/a, which means tan(angle) = "opposite side" over "adjacent side"

In your case, the side $A$ is opposite the angle BC. I recommend you draw a picture to convince yourself that $\tan(70) = \frac{A}{B}$, and therefore $A = (10 {{\rm m}}) \times \tan(70^\circ \times (\frac{\pi {{\rm rad}}}{180^\circ} ) )$.

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  • $\begingroup$ Thanks for the answer Charles +1, So it's PI times radians, where do I get the radians? I'm pretty ignorant I'm sorry $\endgroup$ – Kyle Jan 12 '16 at 16:25
  • $\begingroup$ @Mr.Derpinthoughton Actually, I would recommend you just write $\tan(70^\circ)$ if your instructor is OK with it (and I guess they probably are, since they gave you all the angles in degrees in the first place). If you are supposed to use "pure" trig functions, using only a number of radians as their input (for example, many math libraries for computer programs require this), you can write $\tan\left(70 \times \frac{\pi}{180}\right)$; you might consider $\frac{\pi}{180}$ as a "conversion factor" that converts degrees to radians. $\endgroup$ – David K Jan 12 '16 at 17:18
  • $\begingroup$ I don't have any instructor, I did it at home, I found the angle by using a protractor :) $\endgroup$ – Kyle Jan 12 '16 at 17:23

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