Question

pls answer with explanation, i will mark as brainliest​

Answer 1

Answer :

$\huge\mathfrak\pink{39°}$

Step-by-step explanation:

Given,

ABCD is a rhombus ( AB = BC = CD = AD )

In ∆ACE ,

Angle (ACE + CAE + CEA ) = 180° [ Sum of angles in a ∆ = 180°]

=> Angle ACE + 68° + 33° = 180°

=> Angle ACE + 101° = 180°

=> Angle ACE = 180° - 101°

=> Angle ACE = 79° --------- (¡)

In ∆ADC, AD = CD

So, Angle DAC = Angle ACD = x

=> Angle ( ACD + CAD + CDA ) = 180°

=> x + x + 100° = 180°

=> 2x + 100° = 180°

=> 2x = 180° - 100°

=> 2x = 80°

=> x = 80° / 2 = 40°

=> Angle ACD = Angle DAC = 40° --------(¡¡)

Comparing ∆ABC & ∆ADC,

AC = AC ( Common side )

AB = AD( Side of a rhombus)

BC = CD ( Side of a rhombus)

Therefore , ∆ ABC is congruent to ∆ ADC by S.S.S

=> Angle ABC = Angle ADC = 100° by C.P.C.T.C --------(¡¡¡)

=> Angle BAC = Angle DAC = 40° by C.P.C.T.C-----------(iv)

=> Angle BCA = Angle ACD = 40° by C.P.C.T.C-----------(v)

Since, Angle ACE = Angle BCA + Angle BCE

=> 79° = 40° + Angle BCE

=> 79° - 40° =Angle BCE

=> Angle BCE = 39°

[Note : The angle at which 33° is there is mentioned as Angle E by me]

Hope it helps........✔✔✔

PLEASE FOLLOW ME.......