Question

In the given figure,ABCD is a trapezium. FG is a straight line. Find angle EBF.​

Answer 1

Given :

• ABCD is a Trapezium

• FG is a straight line

To Find :

• Angle EBF

Solution :

In a Trapezium Opposite Sides are Parallel

$\therefore \sf \: AB \parallel \: DC$

When AB and DC are two lines parallel to each other EC is transversal.

$\bullet \: \sf \angle \: BCD = 45^{\circ} \: \: \: (Given)$

When FG is a straight line

$\rightarrow \: \sf \angle \: BCD = \angle \: ABE = 45^{\circ}$

Angle Sum Property :

$\longrightarrow \sf \angle \: ABE + \angle \: GBC+ \angle \: ABG = 180^{\circ} \: \: \: (Linear \: Pair )$

$\longrightarrow \sf 45^{\circ} \: + \angle \: GBC + 30^{\circ} = 180^{\circ}$

$\longrightarrow \sf 75^{\circ} \: + \angle \: GBC = 180^{\circ}$

$\longrightarrow \sf \: \angle \: GBC = 180^{\circ} - 75^{\circ}$

$\longrightarrow \sf \: \angle \: GBC = 105^{\circ} \:$

Now,

$\bigstar \large\boxed{\sf \: \angle \: GBC = \angle \: EBF = 105^{\circ}}$

(Verticallly Opposite Angle)

Hence, Angle EBF = 105°

Answer 2

$\bold{\mathtt{Given}}$

⟹ ABCD is a trapezium

⟹ AB|| DC

⟹ Angle ABG = 30°

⟹ Angle BCD = 45°

$\bold{\mathtt{Solution}}$

⟹ As AB || DC , Angle DCB = Angle ABE

Corresponding angles .

⟹ Angle ABE = 45° eq ( I )

Now Taking linear pair

[ Sum of linear pair is 180° ]

⟹ Angle ABG + ABE + EBF = 180°

Putting value of angle ABG and Angle ABE ( from eq I )

⟹ 30° + 45° + EBF = 180°

⟹ 75° + EBF = 180°

⟹ EBF = 180° - 75°

Angle EBF = 105°