Question

(iv) ZCFB
please mark you Brian list ​

Answer 1

Answer:

CFB=65°

Step-by-step explanation:

first AEB=180°-(90°+20°)=70° ( sum of all sides of triangle=180°)

then DEA=DEF+FEB+BEA=180° ( linear pair )

=45°+FEB+70°=180°

FEB=180°-(45°+70°)=65°

In∆FEB,

FEB+EBF+BFE=180°

65°+EBF+70°=180°

EBF=180°-(65°+70°)=45°

Now, since it is a square, it's angles are of 90° that means CBA=90°=CBF+FBE+EBA

90°=CBF+45°+20°

CBF=90°-(45°+20°)

=25°

In∆CFB,

CFB+CBF+FCB=180°

CFB+25°+90°=180°. (FCB is an angle of square)

CFB=180°-(25°+90°)=65°

Answer 2

Answer:

Follow me back I have followed you

Step-by-step explanation:

Follow =Follow

Unfollow =Unfollow