[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 meng.o3. 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 .
2008-09-09 11:39:57
- meng.o3
- » FTalkElite
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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]
using trigonometric identities, and Angle sum-difference where:
sin (A+B) = sinAcosB + cosAsinB
cos (A+B) = cosAcosB - sinAsinB
sinB + cosB + sin(B+90) + cos(B+180) = 0
sinB + cosB + [b](sinBcos90 + cosBsin90)[/b] + [b](cosBcos180 - sinBsin180)[/b] = 0
[i]sinB[/i] + [u]cosB[/u] + [i]sinBcos90[/i] + [u]cosBsin90[/u] + [u]cosBcos180[/u] -[i] sinBsin180 [/i]= 0
sinB(1 + cos90 - sin180) + cosB(1 + sin90 + cos180) = 0
sinB (1) + cosB(1) = 0
sinB = -cosB [i]divide both sides by cosB[/i]
(sinB)/(cosB) = -(cosB)/(cosB) [i] sin/cos = tan[/i]
tanB = -1
B = tan^1(-1) [i]inverse tangent of -1[/i]
[b]B = -45[/b]
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