...
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?
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 . Click on any LaTex to see the code used.
Now for your problem. Yes, and so that .
Take the "opposite" side of the smaller triangle to be "1" and the "near" side to be . Then the larger triangle has "opposite" side of length 2 and "near" side of length so that . You can find and solve for x from that.
Those will NOT be whole numbers.
This is incorrect. You multiplied when you should have divided.tan (x+30) = L/L under.root (3/4).