What is the slope of a line perpendicular to the line whose equation is 6x+2y=−32. Fully simplify your answer.
1 answer:
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The values of the trigonometry functions are sin(α) = 4/7, cos(β) = 4/7, tan(α) = 4/√33, cot(β) = 4/√33, sec(α) = 7/√33 and sec(β) = 7/√4
<h3>How to evaluate the trigonometry functions?</h3>The figure that completes the question is added as an attachment
From the figure, we have the third side of the triangle to be
Third = √(7^2 - 4^2)
Evaluate
Third = √33
The sin(α) is calculated as:
sin(α) = Opposite/Hypotenuse
This gives
sin(α) = 4/7
The cos(β) is calculated as:
cos(β) = Adjacent/Hypotenuse
This gives
cos(β) = 4/7
The tan(α) is calculated as:
tan(α) = Opposite/Adjacent
This gives
tan(α) = 4/√33
The cot(β) is calculated as:
cot(β) = Adjacent/Opposite
This gives
cot(β) = 4/√33
The sec(α) is calculated as:
sec(α) = Hypotenuse/Adjacent
This gives
sec(α) = 7/√33
The csc(β) is calculated as:
sec(β) = Hypotenuse/Opposite
This gives
sec(β) = 7/√4
Hence, the values of the trigonometry functions are sin(α) = 4/7, cos(β) = 4/7, tan(α) = 4/√33, cot(β) = 4/√33, sec(α) = 7/√33 and sec(β) = 7/√4
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