barker 2001 3 unit solutions (1 Viewer)

[HSC :(]

hai
i was jus wonderin if anyone can do question 2a from this paper

when u do it by calculator u do it right.. but it cant be though cos when u work it out by hand it dusnt turn out right..

thanx

 

[Fosweb]

Draw your triangles to 'work it out by hand'. The acute angle is pi/6 for the cos^-1(-sqrt(3)/2), (easy to see if you use tan^-1(1/-sqrt(3)).
Its negative so the angle is in the opposite quadrant, so it becomes 5pi/6.
Other one is obvious to be pi/4.
5pi/6 - pi/4 = 7pi/12

 

[HSC :(]

a lil prob

one thing though..

if u hav a negative sign for cos, that means that the angle cant be in the first quadrant, so u cant hav pi/4 for the tan, it has to lie in the third quradrant, where tan is positive and cos is negative..

tell me what u think?

 

[Fosweb]

They are separate i think.