Answer:
5√3 = b
5 = d
10√3 = a
15 = c
Step-by-step explanation:
Sin60° = opp/hyp
sin60° = b/10
10sin60° = b
5√3 = b
Cos60° = adj/hyp
Cos60° = d/10
10Cos60° = d
5 = d
Sin30° = opp/hyp
Sin30° = 5√3/a
a = 5√3/Sin30°
a = 10√3
Tan30° = opp/adj
Tan30° = 5√3/c
c = 5√3/tan30°
c = 15
Answer:
18-6=2+y
Step-by-step explanation:
18-6=2+y
X=6 is the answer for the problem
Differentiate both sides of the equation of the circle with respect to
, treating
as a function of
:

This gives the slope of any line tangent to the circle at the point
.
Rewriting the given line in slope-intercept form tells us its slope is

In order for this line to be tangent to the circle, it must intersect the circle at the point
such that

In the equation of the circle, we have

If
, then
, so we omit this case.
If
, then
, as expected. Therefore
is a tangent line to the circle
at the point (1, -2).