VCAA 2010 NON CAS Question

[ashs_vb]

For q3c  I deduced that the triangle was isosceles and hence the height( if we take it from A to AB) is perpendecular to the vector AB. Now if i used this method, Link attached, am i assuming anything? Because i don't get the right answer.
I know that there is an easier way but i was just wondering what i had done wrong.
Thanks

[silverpixeli]

Might be that the cosine from the previous question is the exterior angle since it's the angle between vectors OA and AB when they're lines up tail-to-tail and in the triangle they're not tail-to-tail they're tail-to-head? I remember my diagram was something different to yours.

[ashs_vb]

but arcos(4/9) is acute

[ashs_vb]

but then again it also could be x where cos(theta +x)=0

[Professor Polonsky]

I can't really tell what you did there. Do you mind typing it up?

[ashs_vb]

Its wrong, i used the wrong angle, i could have used cos(x)=-4/9. Its the head to tail issue, you're right.
No worries, and polopotamus, i tried to work out the area by finding the height through the double angle formula as it is perpendecular to the base, then i compared values, like similar triangles. Didn't work though, oh well, back to work!