How do you find the angle #alpha# such that the angle lies in quadrant III and #secalpha=-1.108#?
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"Suppose that I don't have a formula for #g(x)# but I know that #g(1)
= 3# and #g'(x) = sqrt(x^2+15)# for all x. How do I use a linear approximation to estimate #g(0.9)# and #g(1.1)#?"
1 Answer
Feb 18, 2017
Explanation:
cos a = - 0.90 -->calculator gives: arc
Unit circle gives another arc x that has the same cos value:
x = - 154^@16 or
x = 360 - 154.16 = 205^@84.
This arc lies in Quadrant 3.
