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Topic review (newest first)
- austin81
- 2005-11-06 01:52:16
I'll continue from
where the inv of tan:
sin,cos and tan are +ve at 1st quandrant(angle=@°),sin +ve at 2nd(angle=180°-@),tan+ve at 3rd(angle=180°+@°) and cos+ve at 4th(angle=360°-@°).
Invtan(-1/2)=26.6° and Intan(-3°)=71.6°
Therefore required angles are
(180°-26.6)°, (360-26.6)°,(180-71.6)°and(360-71.6)°
Sorry I don't haveagood programthat illustrate Maths symbols.
- Flowers4Carlos
- 2005-11-05 14:18:10
hi yaz lloyd!
i can't remember how to do these exactly, but i think it should go something like this:
2tan²θ + 7tanθ = -3
2tan²θ + 7tanθ + 3 = 0
2(tan²θ + 7/2tanθ + 3/2) = 0
2(tanθ + 1/2)(tanθ + 3) = 0
tanθ = -1/2 tanθ = -3
θ = tan-¹(-1/2) θ = tan-¹(-3)
hmm... i guess u need a calculator to find θ
- Lloyd
- 2005-11-05 11:20:37
hello, i need a little help again on trig. please explain this to me: thanks, lloyd
Solve for θ the quadratic 2tan^2 θ + 7tanθ = -3 where 0<θ<360