When an object is dropped from a building (height 1450 feet )
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If 
is the variable of the horizontal axis, then you can solve for 
to get the equation of the line in slope-intercept form in the 
plane:
i.e. a line with slope
through the origin, which means it is contained in the first and third quadrants. Since the terminal side of 
has a negative sine, the angle must lie in the third quadrant.
Because the slope of the line is, you can choose any length along the line to make up the hypotenuse of a right triangle with reference angle . Any such right triangle will have 
, regardless of whether the angle is the first or third quadrant. But since is known to lie in the third quadrant, and so 
and 
are both negative, you have
i.e. a line with slope
Because the slope of the line is
Let 
. Then 
, and so as 
, you have 
. The limit is then equivalent to
