Answer:
We draw line AB which is perpendicular to the 14 cm side
Since Angle C is 60 degrees that makes angle CAB = 30 degrees
Triangle CAB is a 30 60 90 triangle so line CB is half the hypotenuse or 5 cm
Line BD equals 9 cm
Line AB^2 = 10^2 - 5^2 = 75
Line AD^2 = AB^2 + BD^2
Line AD^2 = 75 + 81
Line AD^2 = 156
Line AD = 12.4899959968
Line AB = Sqr root (75) = 8.6602540378
Angle D = arc sine (8.6602540378 / 12.4899959968)
Angle D = 43.898 degrees
Angle A = 180 - 60 - 43.898 = 76.102 degrees
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Actually this could have been done a little easier by using the Law of Cosines and then the Law of Sines, but I just thought I'd show another way to solve this.
Step-by-step explanation:
(1, 3.5 ) and (3, 3.5 )
the endpoints of the midsegment are at the midpoints of DE and DF
using the midpoint formula with D(0, 7), E(2, 0), F(6, 0)
midpoint of DE = [
(0 + 2),
(7 + 0)] = (1, 3.5)
midpoint of DF = [
(0 + 6),
(7 + 0)] = (3, 3.5)
Y - y 0 = m ( x - x 0 )
x 0 = 6, y 0 = -3, m = 1/2
The point-slope form of the equation is:
y + 3 = 1/2 ( x - 6 )
D. . . . . . . . . . . . . . . . . . . .