Triangle in a circle

[mendhak]

Help! I just figured out I really suck at this.

How can I figure out x (in terms of the angles and other side of course)?

Also, it would be a great help if someone could point out a site to me that teaches this simple stuff.

Thanks.

[SteveCRM]

is that 30 the angle on that side?

if so:

a triangle with 1 side as the diameter is always right. The bottom angle is the right angle. So this makes a 30 60 90 triangle. I don't remember the proportion for any sides...but I think it has sqrt(2) in it.

[prog_tom]

Using Trigonometry:

Given that AY = 15, so:

Y = 30,

X = unknown

if we use the relationship:

sin 30deg = x/30

x = 30 * sin 30deg

x = 15

[kedaman]

assuming that c is a right angle, but its not given.
if ab crosses the circle center, then you could draw a line from it to c and show that it is though.

[sql_lall]

BTW:
sin(30) = 1/2
cos(30) = sqr(3)/2
tan(30) = 1/sqr(3)

Oh, the humiliation if i mixed up 30 and 60 again

[prog_tom]

Originally posted by sql_lall
BTW:
sin(30) = 1/2
cos(30) = sqr(3)/2
tan(30) = 1/sqr(3)

Oh, the humiliation if i mixed up 30 and 60 again

Tan 30deg is sqrt(3)/3, not 1/sqrt(3).

[sql_lall]

wait:

sqr(3)
-------
__3_______(Underlines for spacing reasons)


= 1 *sqr(3)
----------------
sqr(3)*sqr(3)

= 1
------
sqr(3)

[prog_tom]

Still, you were wrong,

denomonators must be rationalized, or it's invalid.

[sql_lall]

fine, fine

yet tan(30) is just an irrational number, and both 1/sqr(3) AND sqr(3)/3 are equal to that number

And, it's actually not 'invalid' , just 'frowned upon' -

[bugzpodder]

denomonators must be rationalized, or it's invalid.

nah its just that teachers like it more that way and they tend to "take marks off" in order to penalise you for not using a "good habit"