I will mark the first answer brainliest so please answer it fast
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since ABCD is a square, show AB = BC = CD = AD
also he is an equilateral triangle, so AD = DE = AE
NOW WE CAN SAY
AB = BC = CD = AD = DE = AE
Now in triangle ABE,
we get AB = AE
which means ABE is an isosceles triangle.
and Angle BAE=angle BAE + angle DAE
→Angle BAE = 90° + 60° = 150°
SINCE ABE is an isosceles triangle,
we can say angle ABE = angle AEB.
SO in Triangle BAE,
angle ABE + angle AEB + angle BAE = 180°
2 × (angle ABE) = 180° - 150° = 30°
angle ABE = angle AEB = 15°.
now find your answer, it's easy now.
also he is an equilateral triangle, so AD = DE = AE
NOW WE CAN SAY
AB = BC = CD = AD = DE = AE
Now in triangle ABE,
we get AB = AE
which means ABE is an isosceles triangle.
and Angle BAE=angle BAE + angle DAE
→Angle BAE = 90° + 60° = 150°
SINCE ABE is an isosceles triangle,
we can say angle ABE = angle AEB.
SO in Triangle BAE,
angle ABE + angle AEB + angle BAE = 180°
2 × (angle ABE) = 180° - 150° = 30°
angle ABE = angle AEB = 15°.
now find your answer, it's easy now.
bblbubbly1:
I just not needed answer instead I need solution
thank you for asking.
if you want solution tell me
ok, I am giving a solution after making
but how is it possible
pls show the soln
ya true 15 is not the final answer, but it's the end.
thank you hope you understand
but we have to find the value of Y
Answered by
1
ABCD is a square, so all angles are 90. A=B=C=D=90.___1). EAD=60. So, In ∆AEB. EAB=EAF+FAB=60+90= 150______2). In ∆AEB, A+E+B=180. 150+x+x=180; we get, x=15. In ∆AED, E=60. AED=AEF+FED ; 60= 15+FED. so, FED=45. In ∆FED, D=60. F+E+D=180. so, F+60+45= 180. we get, F= 75. F=y =75
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