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 $\sqrt{x}$. Click on any LaTex to see the code used.

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

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