well im solving for t
2cos t = 6cos t - squreroot 12
3cos t= -sq rt 12
cos t = -sqr rt 12/3
cost t=2 sq rt 3/3
the angle is 30 how do i get it???
2007-12-27 15:37:53 · 6 answers · asked by suzy 1 in Science & Mathematics ➔ Mathematics
2 cos t = 6 cos t - √12
√12 = 4 cos t
2√3 = 4 cos t
(√3) / 2 = cos t
To solve for t using a calculator, take the inverse cosine (cos‾¹) of "(√3) / 2". You will get t = 30°
You can usually find the inverse cosine button by pressing the "2nd" or "shift" button on your scientific calculator, then the "cos" button.
You could also use a table of cosines (or you should be familiar with √3 and 2 being sides of a 30-60-90 triangle) to find your answer
2007-12-27 15:48:17 · answer #1 · answered by lcamccandlj 3 · 0⤊ 0⤋
a) 2cos t = 6cos t - squareroot 12
2cos t - 6 cos t = - sqrt(12)
- 4 cos t = - sqrt(12)
cos t = - sqrt(12)/-4
t = arccos[sqrt(12)/4]
t = 30 ANS
b) 3cos t = - sqrt(12/3)
3cos t = -sqrt(4)
cos t = -2/3
t = arcco(-2/3)
t = 131.81
c) cos t = - sqrt(12/3)
cos t = - 2
t = arccos(-2)
t = undefined
d) cos t = 2 sqrt(3/3)
cos t = 2(1)
t = arccos(2)
t = undefined
teddy boy
2007-12-27 23:53:41 · answer #2 · answered by teddy boy 6 · 0⤊ 1⤋
but the steps you have is not correct
4cos t = sqrt(12)
cos t= 2sqrt(3)/4
cos t = sqrt3/2
t = 30 deg
about using arc cos or cos-1, usually it is a shift fn on most calc.
2007-12-27 23:51:00 · answer #3 · answered by norman 7 · 0⤊ 0⤋
2cost = 6cost - sqrt(12)
subtract 6cost from both sides
-4cost = -sqrt(12)
cost = (sqrt(12))/4
= 0.866025
t = acos(0.866025)
= 0.523599583 radians
= 30 degrees
2007-12-27 23:57:12 · answer #4 · answered by pico t 2 · 0⤊ 0⤋
use
arccos(2 sqrt(3)/3)
it is also cos^-1 on many calculators
2007-12-27 23:42:51 · answer #5 · answered by Mαtt 6 · 0⤊ 1⤋
depends on what calculator you got, what you have?
2007-12-27 23:43:07 · answer #6 · answered by Patrick 3 · 0⤊ 0⤋