The point is in the 1st quadrant, so all our ratios will be positive
r^2 = 63^ + 280^2 = 82369
r =√82369 = 287
sinA = y/r = 280/287 = 40/41 , then cscA = 41/40
cosA = x/r = 63/287 = 9/41 , then secA = 41/9
tanA = y/x = 280/63 = 40/9 , then cotA = 9/40
Find the values of the six trigonometric functions for ∠A in standard position determined by the point (63, 280). Enter your answers as fractions.
1 answer