# Finding the apex - Simple Trig, but can't figure it out.

• Oct 11th 2010, 09:27 AM
bkelly301
Finding the apex - Simple Trig, but can't figure it out.
Attachment 19274

In the above drawing, it is simple to find the angles and legs of each individual triangle using Trig. For example, the one angle is Ø = ATan(((3.0 - 2.5)/2)/2.1).

The other angle is, Ø = ATan((((3.0 - 2.6)/2)/(1.6 - 0.9)).

Using those angles, I can find the hypotenuse of either triangle. However, in order to find "X" in this drawing, it is dependant on BOTH angles, and not just one.

Usually in Trig, all I need is one angle and one side, or 2 sides.....but when something like this, which involves 2 angles....I'm stumped!!!

Does anyone know how to solve this?

Thanks a lot!
• Oct 13th 2010, 10:59 AM
earboth
Quote:

Originally Posted by bkelly301
Attachment 19274

In the above drawing, it is simple to find the angles and legs of each individual triangle using Trig. For example, the one angle is Ø = ATan(((3.0 - 2.5)/2)/2.1).

The other angle is, Ø = ATan((((3.0 - 2.6)/2)/(1.6 - 0.9)).

Using those angles, I can find the hypotenuse of either triangle. However, in order to find "X" in this drawing, it is dependant on BOTH angles, and not just one.

Usually in Trig, all I need is one angle and one side, or 2 sides.....but when something like this, which involves 2 angles....I'm stumped!!!

Does anyone know how to solve this?

Thanks a lot!

I can show you a way to solve this question, but I don't use trigonometry.

1. I define a coordinate system. The axes are drawn in blue.

2. At the left side of the image you can determine 4 points by their coordinates:

A(0, 2.5), B(0.25, 0.4). Then the straight line through A and B has the equation:

$\displaystyle y-2.5 = \dfrac{-2.1}{0.25} \cdot x~\implies~y=-8.4x+2.5$

C(0, 0), D(0.2, 1.6). Then the straight line Through C and D has the equation:

$\displaystyle y = \dfrac{1.6}{0.2} \cdot x ~\implies~y=8x$

3. Calculate the coordinates of the point of intersection. You'll get $\displaystyle I\left(\frac{25}{164} \ ,\ \frac{50}{41}\right)$

4. Do the same at the right side of the image with the points A', B', C', D' and I'.
Calculate the length $\displaystyle X=|\overline{II'}|$. I didn't label the points!

5. I've got $\displaystyle |\overline{II'}| =X= \dfrac{221}{82}$
• Oct 13th 2010, 12:03 PM
bkelly301
Wow, I would have never thought about doing it that way. Thanks a lot! At my job, I use trig all the time, but I haven't used basic y=mx + b algebra since I was in school, so I forgot all about that.

The whole point of this is that I want to make up a little excel program where I plug in all the given values (all the numbers shown), and it outputs what X will be. Now that I know how to solve this problem mathematically, I can figure out how to put it into an excel program.

Much appreciated. Thanks again!