Can someone explain how I can triangulate using angles and one side of a Right-angled triangle?

Question

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

Answer 1

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

Answer 2

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 $A = (10 {{\rm m}}) \times \tan(70^\circ \times (\frac{\pi {{\rm rad}}}{180^\circ} ) )$.