The question says that angle CED = theta, but the diagram says that angle CED = q.no i haven't.
Angle(ACB) as u've indicated in ur diagramwhere is the right angle for part ii?
I guided with the help of diagram and i used to brain powerYer, I'm not derping, angle ACB can't be 90.
Why not? Sorry, I'm just confused as to why it can't be 90.Yer, I'm not derping, angle ACB can't be 90.
GOOD POINT!But you cannot assume that angle ACB is a right angle. That would be too easy for 3 marks. Since part ii) is 3 marks, you probably have to show that triangle ABC has a right angle and then use SOHCAHTOA to find sin(theta).
Assume ACB is 90, and since angle EDC Is 90, then . But this isn't possible as and so the two lines BC and DE shouldn't intersect. But they do intersect (at D), so our assumption must be wrong, that is ACB isn't 90.Why not? Sorry, I'm just confused as to why it can't be 90.
That's cos it's not.wowowowo
Angle(BAC) = theta?
how?
Yep this.it isnt, the markings on the diagram are wrong.
i repeat cos rule anyone?