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secondary 3 | A Maths
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Explain if possible and workings pls
Date Posted: 3 years ago
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130 is an angle in the second quadrant, with basic angle 180 - 130 = 50
(I can’t draw a diagram in this chat)
What this means is that cos 50 and cos 130 have numbers which are numerically equal (without considering signages).
However, we know that 130 is an angle in the second quadrant whereas 50 is an angle in the first quadrant, so cos 130 will have a negative value whereas cos 50 will have a positive value. So, the signage of cos 130 will be opposite that of cos 50.
Since cos 50 = a, we have
cos 130 = -a
You might have seen this in E Maths under the supplementary angle rule before.
(I can’t draw a diagram in this chat)
What this means is that cos 50 and cos 130 have numbers which are numerically equal (without considering signages).
However, we know that 130 is an angle in the second quadrant whereas 50 is an angle in the first quadrant, so cos 130 will have a negative value whereas cos 50 will have a positive value. So, the signage of cos 130 will be opposite that of cos 50.
Since cos 50 = a, we have
cos 130 = -a
You might have seen this in E Maths under the supplementary angle rule before.
See 1 Answer
Cos(180-130) degree is cos(50 degree)
Any trigometry (theta) = +or - Any trigometry (180 x n + or - theta)
Remember
All function is + in first Quandrant
Cosine is plus in forth Tangwbtviscthird Sine is second
ACTS
Therefore COS(180-130) is - COS(130) = COS50 degree = a
Therefore -a
Got it?
Go
Any trigometry (theta) = +or - Any trigometry (180 x n + or - theta)
Remember
All function is + in first Quandrant
Cosine is plus in forth Tangwbtviscthird Sine is second
ACTS
Therefore COS(180-130) is - COS(130) = COS50 degree = a
Therefore -a
Got it?
Go
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