Given that cos Ф = 5/6 I know that one side is 5 and the hypotenuse is 6 so the other side is:
a^2 = c^2 - b^2
a^2 = 6^2 - 5^2
a^2 = 36 - 25 = 11
a =
Now I know that the sin is
since sin is defined as opposite/hypotenuse
the second one uses a form of C = pi * d we are going to add the sector part to it.
C = 6 and the central angle is 4 degrees we can set up like this:
d = 171.974
r = 171.94/2 =85.987
Cot(5pi/12) = tan(pi/12)
tan(x/2) = (1 - cos(x)) / sin(x))
tan(pi/12) = tan((pi/6)/2) = (1 - cos(pi/6)) / sin(pi/6))
sin(pi/6) = 1/2
cos(pi/6) = sqrt(3) / 2
so...
tan(pi/12) = (1 - sqrt(3)/2) / (1/2) = 2 - sqrt(3)
I am not sure, but i think its root of 3 bcs tan of the angle given is the slope
Answer:
a^4 + a^2 + 1
a^4 + a^2 + x + 1
a^2 ( a^2 + 1) + 1 ( x + 1)
so, here clearly we can see that If wee add a^2 in place of x it would be a perfect square..
Hope it helpz~