Answer:
A = 50°
B = 60°
C = 70°
Step-by-step explanation:
If we draw a line from each vertex through the center of the circle, we perpendicularly bisect the line joining the adjacent tangent points.
We then know the original angle is halved and the remaining angle of each right triangle is complementary to half the original.
Now we can subtract the known angles along each line of the original side to find the remaining angle
They both have the same rate of change
35
since m and n are the midpoints of ab and ac resistively
then ,bc=2mn=16
since m and l are the midpoints of ba and bc respectively
then, ac =2mn =10
then,nc =1/2 × 10 =5
similary, mb=6
then ,perimeter=nc+mb+mn+bc=5+6+8+16=35
First one:
cos(A)=AC/AB=3/4.24
cos(B)=BC/AB=3/4.24
Cos(A)/cos(B)=AC/AB / (BC/AB) = AC/AB * AB/BC = AC/BC=3/3=1
Second one:
To solve this problem, we have to ASSUME AFE is a straight line, i.e. angle EFB is 90 degrees. (this is not explicitly given).
If that's the case, AE is a transversal of parallel lines AB and DE.
And Angle A is congruent to angle E (alternate interior angles).
Therefore sin(A)=sin(E)=0.5
Answer:
even I think
Step-by-step explanation:
