m∠FDE = 52°
Solution:
Given data:
DE ≅ DF, CD || BE, BC || FD and m∠ABF = 116°
<em>Sum of the adjacent angles in a straight line = 180°</em>
m∠ABF + m∠CBF = 180°
116° + m∠CBF = 180°
m∠CBF = 64°
If CD || BE, then CD || BF.
Hence CD || BE and BE || FD.
Therefore BFCD is a parallelogam.
<em>In parallelogram, Adjacent angles form a linear pair.</em>
m∠CBF + m∠BFD = 180°
64° + m∠BFD = 180°
m∠BFD = 116°
<em>Sum of the adjacent angles in a straight line = 180°</em>
m∠BFD + m∠DFE = 180°
116° + m∠DFE = 180°
m∠DFE = 64°
we know that DE ≅ DF.
<em>In triangle, angles opposite to equal sides are equal.</em>
m∠DFE = m∠DEF
m∠DEF = 64°
<em>sum of all the angles of a triangle = 180°</em>
m∠DFE + m∠DEF + m∠FDE = 180°
64° + 64° + m∠FDE = 180°
m∠FDE = 52°
Answer:
c
Step-by-step explanation:
the right answer is c it good
The answer is d>3
I cant put it graphed but it is open circle on three and the line pointing 4, 5, 6, 7,
Using the quadratic formula
Answer:
y=-5/6x-5
Step-by-step explanation:
0-(-5)/-6-0
-5/6 slope
y-(-5)=-5/6(x-0)
y+5=-5/6x
y=-5/6x-5
