Can someone pleasehelp me with thses math questions?

Question

there are only two please help me. :)

http://i230.photobucket.com/albums/ee247/memorex456/scan0002.jpg

http://i230.photobucket.com/albums/ee247/memorex456/scan0001.jpg

Answer 1

let the center of the circle be c

the angle xcy = 360/3 =120 degrees

angle cxy = (180 -120)/2 = 60/2 =30 (isoceles triangle)


cos(30) = (w/2)/4 = w/8 ----> w = 8 cos(30)
cos(30) =(√3 ) /2

w = 8 √3 /2 = 4 √3

********************
(2 -2 sin²x )/cos x = 2(1- sin² x)/cos x

use the identity:
cos² x +sin² x =1 or
cos² x = 1 -sin² x

2(1- sin² x)/cos x = 2(cos² x)/cos x = 2 cos x

Answer 2

Hello,

First one. Since the circle is divided into three equal angles they are each 120 degrees. So if we draw an altitude to the bottom we have two 30, 60, 90 triangles then in the left triangle we have the hypotenuse to be 4 and the altitude is 1/2 this or 2 and the base is srt3 times the short side or 2sqrt3 now since this is half the base then we double it an get 4 sqrt3 or 4*1.732 = 6.928

Hope This Helps!!

Answer 3

Let z the center of circle.

Angle xzy = 120°
Triangle xzy = isosceles

Angle wxz = 30°
cos30° = xw/4

xw = 3.46
xy = 6.93

============================


2 (1 - sin^2x) / cosx = 2cos^2x / cosx = 2 cosx

Answer 4

First one is 6.92 (Rounded)

Second one, I don't know because I haven't learned Cosine yet ...

Answer 5

2 - 2sin^2(X ) / cosX

= 2(1 - sin^2(X)) / cosX

= 2cos^2(X) / cosX

= 2cosX