We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
127
2
avatar

A regular hexagon has vertices at $1-i, 0, a, b, c, d$ clockwise in that order, as in the picture below:

Then what is \(\operatorname{Im}(a)\)?

 Jan 16, 2019
 #1
avatar+533 
0

if you tell me what Im(a) means(does it mean imaginary part?), i can solve it, but i can tell you the coordinates.

 

convert to the real plane, so distance from a to origin is sqrt2/2. now since the angle made by a, origin, and y axis is 15 degrees, you use the fact that cos 15 = (sqrt6+sqrt2)/4, and the fact that sin 15 = (sqrt6-sqrt2)/4, so then the coordinates of a would be ((sqrt6-sqrt2)/8, (sqrt6+sqrt2)/8).

 

assuming it means imaginary part, answer is (sqrt6+sqrt2)/8.

 Jan 16, 2019
 #2
avatar+21984 
+9

A regular hexagon has vertices at $1-i, 0, a, b, c, d$ clockwise in that order, as in the picture below:

 

Then what is \(\operatorname{Im}(a)\)?

\(\begin{array}{|rcll|} \hline \sin(75^{\circ}) &=& \dfrac{\operatorname{Im}(a)} {\sqrt{2}} \\ \operatorname{Im}(a) &=& \sqrt{2}\cdot \sin(75^{\circ}) \\ \operatorname{Im}(a) &=& 1.41421356237\cdot 0.96592582629 \\ \mathbf{\operatorname{Im}(a)} & \mathbf{=} & \mathbf{1.36602540378} \\ \hline \end{array}\)

 

laugh

 Jan 16, 2019

10 Online Users

avatar
avatar