Question

please answer this question in properly way ​

Answer 1

$\huge {\mathbb {\green {\boxed {ANSWER}}}}$

$\tiny {\green {\underline {*⟨\: represents \:the\: symbol \: of\: angle\: here.}}}$

$\large {\mathtt {\green {FIRST \: CASE}}}}$

If BE is extended to DC, to a point F, then

⟨ F = ⟨ B = 35° (Interior Alternate Angles)

We know that all the angles of a triangle sum up to 180°.

So, in ∆FED

=> 180° - ( ⟨ F + ⟨ D) = ⟨ E

=> 180° - (35° + 65°) = ⟨E

=> 180° - 100° = ⟨ E

=> 80° = ⟨ E

Angle x is adjacent to ⟨ E.

Thus, x = 180° - ⟨ E

= 180° - 80°

= 100°

x = 100° (Ans)

$\large {\mathtt {\green {SECOND \: CASE}}}}$

If BE is extended to DC, to a point F, then

⟨ F = ⟨ B = 55° (Interior Alternate Angles)

We know that all the angles of a triangle sum up to 180°.

So, in ∆FED

=> 180° - ( ⟨ F + ⟨ D) = ⟨ E

=> 180° - (55° + 25°) = ⟨E

=> 180° - 80° = ⟨ E

=> 100° = ⟨ E

Angle E1 is adjacent to ⟨ E.

Thus, E1 = 180° - ⟨ E

= 180° - 100°

= 80°

E1 = 80° (Ans)

Therefore, x = 360° - ⟨ E1

= 360° - 80°

= 280° [Answer— since it's a reflex angle]

$\large {\mathtt {\green {THIRD \: CASE}}}}$

If AE and DC are extended such that they intersect a point F, then

⟨ F = ⟨ A = 116° (Exterior Alternate Angles)

⟨ F1 = 180° - ⟨ F

= 180° - 116°

= 64°

⟨ C1 = 180° - ⟨C

= 180° - 124°

= 56°

We know that all the angles of a triangle sum up to 180°.

So, in ∆FEC

=> 180° - ( ⟨ F1 + ⟨ C1) = ⟨ E

=> 180° - (64° + 56°) = ⟨E

=> 180° - 120° = ⟨ E

=> 60° = ⟨ E

Angle E is adjacent to x

Thus, x = 180° - ⟨ E

= 180° - 60°

= 120°

x = 120° (Ans)

For procedure, refer attachments. There, 'C' represents 'Case', as 'C1' will be 'Case 1'.

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