Question

hi friends
plz answer​

Answer 1

$\bold{GIVEN}$

Line AB and CD are parallel ,

So,

$\mathsf{\angle A = 36}$

Reason :-

Alternate interior angles

Now, in ΔECD

$\mathsf{\angle ECD = 36^{\circ}}$

$\mathsf{\angle EDC= 32^{\circ}}$

By angle sum property sum of all angles in triangle is equal to 180 degress.

So,

$\angle EDC +\angle ECD +\angle CED = 180$

$\mathsf{32 +36 +\angle CED = 180}$

$\mathsf{68 +\angle CED = 180}$

$\mathsf{\angle CED = 180 - 68}$

$\mathsf{\angle CED = 112^{\circ}}$

By straight line angle property,

$\mathsf{\angle CED + \angle ACE = 180}$

Since,

$\mathsf{\angle CED = 112^{\circ}}$

so,

$\mathsf{112 + \angle ACE = 180}$

$\mathsf{\angle ACE = 180 -112}$

$\mathsf{ \angle ACE = 68^{\circ}}$

Since, CE = AC ,given in question so,

$\mathsf{\angle ACE = \angle b = 68^{\circ}}$

Now, in ΔACE

$\mathsf{\angle c +\angle b +\angle CED = 180}$

$\mathsf{\angle c +68 +68= 180^{\circ}}$

$\mathsf{\angle c +136 = 180^{\circ}}$

$\mathsf{\angle c = 180-136}$

$\mathsf{\angle c = 44^{\circ}}$

So,

$\large \boxed{\mathsf{\angle a= 36^{\circ}}}$

$\large \boxed{\mathsf{\angle b= 68^{\circ}}}$

$\large \boxed{\mathsf{\angle c= 44^{\circ}}}$

Hence,

$\bold{OPTION \: 2 \: IS \: CORRECT}$

Answer 2

Solution:-

To Find:-

a = ?

a = ?b =?

a = ?b =?c = ?

Find:-

In ∆ ECD,

∠ECD + ∠EDC + ∠CED = 180° (By Angle Sum Property of a Triangle)

=) 32° + 36° + ∠CED = 180°

=) ∠CED = 180° - 68°

=) ∠CED = 112°

Now,

∠CED + ∠CEA = 180° ( Linear Pair)

=) ∠CEA = 180° - 112°

=) ∠CEA = 68°

Now, In Isosceles ∆ AEC

AC = EC

=) ∠CAE = ∠CEA

=) ∠CAE = b = 68°

In ∆ ACE,

∠ACE + ∠CAE + ∠CEA = 180° ( A.S.P.)

=) c = 180° - 68° - 68°

=) c = 44°

And

Since, AB || DC

=) ∠a = ∠ADC ( By Alternate Interior Angle)

=) ∠a = 36°.

Hence,

∠a = 36°

∠a = 36°∠b = 68°

∠a = 36°∠b = 68°∠c = 44°