the slope of the line passing through the points (2,sintheeta) , (1,costheeta)is 0 then general solution of theeta
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2
Slope = (y2-y1)/(x2-x1)
Here y1=Sinø, y2= Cosø
x1=2,x2= 1
Putting in above equation we get
(Sin ø-Cosø)/(1-2)=0
Sin Ø= Cos Ø
Sinø= Sin( 90°-ø)
ø= 90°-ø
so ø=45°
Here y1=Sinø, y2= Cosø
x1=2,x2= 1
Putting in above equation we get
(Sin ø-Cosø)/(1-2)=0
Sin Ø= Cos Ø
Sinø= Sin( 90°-ø)
ø= 90°-ø
so ø=45°
Answered by
1
Slope(m) of a line passing through the points(x1,y1) and (x2,y2)
m= (y2-y1)/(x2-x1)
(cosθ-sinθ)/(1-2) =0
Cosθ-sinθ=0
cosθ=sinθ
Therefore
θ=45 degrees
Since cos45=sin45
m= (y2-y1)/(x2-x1)
(cosθ-sinθ)/(1-2) =0
Cosθ-sinθ=0
cosθ=sinθ
Therefore
θ=45 degrees
Since cos45=sin45
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