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!!

:)

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- Nov 17th 2011, 12:07 AMJesseElFantasmaGrade 8 Geometry
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, 12:14 AMQuackyRe: Grade 8 Geometry
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 $\displaystyle 360^o$

-Angles in an equilateral triangle are all $\displaystyle 60^o$

-Angles in a square are all $\displaystyle 90^o$

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 $\displaystyle 180^o$. - Nov 17th 2011, 12:14 AMProve ItRe: Grade 8 Geometry
- Nov 17th 2011, 12:23 AMJesseElFantasmaRe: Grade 8 Geometry
Thanks! This helped alot.

So from what you said I drew the conclusion of:

$\displaystyle 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, 12:27 AMJesseElFantasmaRe: Grade 8 Geometry
Thanks! This helped alot.

So from what you said I drew the conclusion of:

$\displaystyle 360^o - (90^o + 60^o + 60^o )= 360^o - 210^o= 150^o...................180^o - 150^o = 30^o........................30^o/ 2....= 15^o................x= 15^o$ - Nov 17th 2011, 12:28 AMQuackyRe: Grade 8 Geometry
Perfect.