[quote=sugarstyx_x]:idea: PLEASE READ: • You're not allowed to Plagiarize, we do not encourage this. • Members can help you but not allowed to do the work for you. • This is a Discussion thread
You are viewing a post by ~jps~. View all 386 posts in [quote=sugarstyx_x]:idea: PLEASE READ: • You're not allowed to Plagiarize, we do not encourage this. • Members can help you but not allowed to do the work for you. • This is a Discussion thread .
2008-09-14 06:07:49
- ~jps~
- » n00b
84
0
1969-12-31
Re: [quote=sugarstyx_x]:idea: PLEASE READ: • You're not allowed to Plagiarize, we do not encourage this. • Members can help you but not allowed to do the work for you. • This is a Discussion thread
[quote=angpogimo]please...anyone!!help me!!
Find the angle of B(assume it as Beta):
sinB + cosB + sin(B+90º) + cos(B+180º) = 0
(show Solution please)
thanks!!![/quote]
another solution: using complex numbers .... convert either cosine or sine in terms of each other
since 1<n = sin (B+n)
therefore all cosine functions must be converted into sine functions
sin W = cos (W+90)
cos W = -sin (w+180)
then add them
sinB + cosB + sin (B+90) + cos (B+180) = 0
= sinB - sin (B+180) + sin (B+90) - sinB = 0
= 1<0 - 1<180 + 1<90 - 1<0 = 0
then the results is 1.414<45
square root of 2 times sin (B+45) = 0
arc sin 0 = (B+45)
0 = B + 45
[b]
B = -45 [/b]
It is very useful when you r dealing with AC circuits analysis
Last edited by ~jps~ (2008-09-14 06:11:27)
 |