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Section 4.4
Question 9b
Question 9b
I can drop a perpendicular from the point (-8, 15) to the line y = 0 to form a right triangle. The given point gives me the value of x and y. To find the hypotenuse, I use a^2 + b^2 = c^2. Doing so, I found the hypotenuse to be 17.
Note: Must know where the trig functions are positive and negative on the xy-plane.
Answer:
sin (theta) = 15/17
cos (theta) = -8/17
tan (theta) = 15/8
csc (theta) = 17/15
sec (theta) = -17/8
cot (theta) = 8/15
Yes?
Question 9b
Question 9b
I can drop a perpendicular from the point (-8, 15) to the line y = 0 to form a right triangle. The given point gives me the value of x and y. To find the hypotenuse, I use a^2 + b^2 = c^2. Doing so, I found the hypotenuse to be 17.
Note: Must know where the trig functions are positive and negative on the xy-plane.
Answer:
sin (theta) = 15/17
cos (theta) = -8/17
tan (theta) = 15/8
csc (theta) = 17/15
sec (theta) = -17/8
cot (theta) = 8/15
Yes?