Posted by Anonymous on Wednesday, November 27, 2013 at 11:42pm.
If cot(2Ø) = 5/12 , then
tan(2Ø) = 12/5 and 2Ø is in either I or III
tan 2Ø = 2tanØ/(1 - tan^2 Ø)
let tanØ = u
then 12/5 = 2u/(1 - u^2) , (for easier typing)
10u = 12 - 12u^2
12u^2 + 10u - 12 = 0
6u^2 + 5u - 6 = 0
(2u + 3)(3u - 2) = 0
u = -3/2 or u = 2/3
tanØ = -3/2 or tan u = 2/3
case 1:
tanØ = -3/2--> Ø in II or IV
make a sketch of a right right-angled triangle
hypotenuse = √(3^2+2^2) = √13
IN II: sinØ = 2/√13 , cosØ = -3/√13, tanØ = -3/2
IN IV: sinØ = -2/√13 , cosØ = 3/√13, tanØ = -3/2
case 2:
tanØ = 2/3---> Ø in I or III
make a new sketch, hypotenuse is still √13
IN I: sinØ = 3/√13, cosØ = 2/√13 , tanØ = 2/3
IN III: sinØ = -3/√13 , cosØ = -2/√13, tanØ = 2/3
check my arithmetic, should have written it out on paper first.