Asked by Tebatso

Determine B e(-360;360)
Cos(B+80)=sin(-300)cos45÷cos405
Answered by mathhelper
preliminary:
sin(-300)
= -sin(300) = -sin(-60) = sin60 = √3/2

cos 45 = √2/2
cos405 = cos45 = √2/2

Cos(B+80)=sin(-300)cos45÷cos405
= √3/2 * √2/2 / (√2/2) = √3/2

then
B + 80° = 30° , B + 80° = 330°, B + 80° = -330°, B+80° = -30°

B = ± 30°, ± 330°
Answered by oobleck
Care to revise your final line?
Answered by mathhelper
Arggghhh,
forgot to subtract 80° from each of these.
so
B = ± 30°- 80°, ± 330° - 80°
= -50°, -110°, 250°, 310°

thanks oobleck
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