Even if you can answer one it would be tremendously appreciated.
Answers
Step-by-step explanation:
first question
angle ABE+angleABD+angleDBC=180°(linear pair)
90°+3x+7+5x+11=180°
8x+108°=180°
8x=180-108
8x=72°
x=72/8
x=9
angleABD=3×9+7=27+7=34°
angleDBC=5×9+11=45+11=56°
angleDBC+angleFBC=180°
56°+3y+1=180°
3y+57°=180°
3y=180°-57°
3y=123°
y=123/3
y=41
angle FBC =3×41+1=123+1=124°
angleEBF+angle FBC=180°
x+124°=180°
x=180°-124°
x=56°
angle EBF =56°
Answer:
1) in angle ABC
3X+7 + 5X+11=90°
SO 8X +18=90
72= 8X
X=9
NOW ANGLE EBF+ANGLE FBC =180°
SO EBF = 180-3Y -1
FOR FINDING Y IN LINE FBD
5X+11+3Y +1= 180
5X+3Y+12= 180
AS X= 9
45+12+3Y=180
3Y= 123
Y=41
so ebf = 56°
2) bca = 44 ( using property )
total sum of angle in a quadrilateral is 360°
adding ang agb + gbc + bca +cag = 360°
90+90+104+x+44= 360
x= 34
3)
2x+3 and 4x-5 are opposite anges so they will be equal soo
2x+3= 4x-5
2x= 8
x=4
now in right angle
4x-5+ ang abc = 90°
ang abc = 90-4(4)+5
= 95-16
= 79