# Thread: Trigno problem. Expressing values in terms of and showing...

1. ## Trigno problem. Expressing values in terms of and showing...

Me again.

Okay, for this question that I'm having difficulty in, here is the figure:

And as we know:

sin30= under.root (1/4)
cos30= under.root (3/4)

This question is divided into 2 parts.

i) express each of AC and BC exactly in terms of L, and show that
AB=(1/2)L under.root7. (Note: how do I write mathematical symbols in this forum?)

I used pythagoras method and sin30=perpendicualr/hypotenuse to get values of DC and AC, but the problem is, they are coming as fraction under.root values rather then whole numbers. This wouldn't be a problem if there weren't any conditions to have under.root solutions. E.g, I'm getting AC as L under.root 3/4. I then have to apply THIS to get DC and subsequently, BC (twice DC), from pythag method. DC comes as (1/2)L therefore BC becomes L. Is this correct?

ii) Even though I was able to prove AB as (1/2)L under.root7, when applying in this part of the question, it doesn't work. Here is says: Show that x=tan^-1(2/under.root3) -30.

I'm doing this like this: tan (x+30) = BC/AC

tan (x+30) = L/L under.root (3/4).
x = tan^-1 [L/L under.root (3/4)] -30

this is wrong but I do not know how to solve this. How do I do this?

2. ...

3. you put maths in a post with 'latex' tags. you should read the latex help forum. and if you click on a latex equation in a post the code pops up for you also. I will let someone else answer your question I am to new to give you the correct advice.

4. This forum uses [ math ] and [ /math ] (without the spaces) to begin and end LaTex. the code for "root" is \sqrt{x} so, for example, [ math ]\sqrt{x}[ /math ] (again without the spaces) gives $\displaystyle \sqrt{x}$. Click on any LaTex to see the code used.

Now for your problem. Yes, $\displaystyle sin(30)= \frac{1}{2}$ and $\displaystyle cos(30)= \frac{\sqrt{3}}{2}$ so that $\displaystyle tan(30)= \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}= \frac{1}{\sqrt{3}}$.

Take the "opposite" side of the smaller triangle to be "1" and the "near" side to be $\displaystyle \sqrt{3}$. Then the larger triangle has "opposite" side of length 2 and "near" side of length $\displaystyle \sqrt{3}$ so that $\displaystyle tan(x+ 30)= \frac{2}{\sqrt{3}}$ . You can find $\displaystyle x+ 30= arctan(\frac{2}{\sqrt{3}})$ and solve for x from that.

Those will NOT be whole numbers.

tan (x+30) = L/L under.root (3/4).
This is incorrect. You multiplied when you should have divided.

5. Thanks for that detailed explanation.

I had completely forgotten that under.root(3/4) could be written as (under.root 3)/2. That clears it. Problem solved.