• Nov 17th 2011, 01:07 AM
JesseElFantasma
Hii!

I'm a bit stuck on this problem which I have. It tells us to find the angle marked x . I tried my best to copy this diagram into paint, so it seems close to the real drawing to me.

http://i44.tinypic.com/fuo0mv.jpg

Thanks!!

:)
• Nov 17th 2011, 01:14 AM
Quacky
We have a square, two equilateral triangles and an isosceles triangle. Focus on working out the largest angle in the isosceles triangle. You can do this using the facts that:
-Angles at a point sum $360^o$
-Angles in an equilateral triangle are all $60^o$
-Angles in a square are all $90^o$

Once you've worked out the largest angle in the isosceles triangle, use the facts that the other two angles in the triangle are equal and that all angles in any triangle sum $180^o$.
• Nov 17th 2011, 01:14 AM
Prove It
Quote:

Originally Posted by JesseElFantasma
Hii!

I'm a bit stuck on this problem which I have. It tells us to find the angle marked x . I tried my best to copy this diagram into paint, so it seems close to the real drawing to me.

http://i44.tinypic.com/fuo0mv.jpg

Thanks!!

:)

What is the size of the angles in a square? What is the size of the angles in an equilateral triangle?
• Nov 17th 2011, 01:23 AM
JesseElFantasma
Thanks! This helped alot.
So from what you said I drew the conclusion of:

$360^o - (90^o + 60^o + 60^o )
= 360^o - 210^o
= 150^o

180^o - 150^o
= 30^o

30^o / 2
= 15^o

x= 15^o$
• Nov 17th 2011, 01:27 AM
JesseElFantasma
$360^o - (90^o + 60^o + 60^o )= 360^o - 210^o= 150^o...................180^o - 150^o = 30^o........................30^o/ 2....= 15^o................x= 15^o$