Question

If ABCD is a cyclic trapezium in which AD∥ BC and ∠B=60o, then ∠BCD is equal to *

A) 80o
B) 90o
C) 60o
D) 100o​

Answer 1

SOLUTION

TO CHOOSE THE CORRECT OPTION

If ABCD is a cyclic trapezium in which AD ∥ BC and ∠B=60°, then ∠BCD is equal to

A) 80°

B) 90°

C) 60°

D) 100°

EVALUATION

Here it is given that in a cyclic trapezium ABCD ∠B = 60°

∴ ∠ABC = 60°

We are aware of the formula the sum of the opposite angles of a cyclic trapezium = 180°

So ∠ABC + ∠ADC = 180°

⇒ 60° + ∠ADC = 180°

⇒ ∠ADC = 180° – 60°

⇒ ∠ADC = 120°

Again it is given that AD || BC

∴ ∠ADC + ∠BCD = 180°

⇒ 120° + ∠BCD = 180°

⇒ ∠BCD = 180° – 120° = 60°

FINAL ANSWER

Hence the correct option is C) 60°

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Answer 2

$\purple\bigstar$Answer:

• (C) 60°

$\green\bigstar$ Given:

• ABCD is a cyclic trapezium.
• In cycle trapezium ABCD, AD || BC
• $\angle \sf \: ABC \: = \: { \sf60}^{ \circ}$

$\pink\bigstar$To find:

• $\angle \: \sf \: BCD$

$\blue\bigstar$ Solution:

$\pink\star$Method 1 :

A quadrilateral is called a cyclic quadrilateral if all the four vertices of it lie on a circle.

So, the vertices of the cyclic trapezium will lie on the circle.

$\because$ We know that the sum of either pair of opposite angles of a cyclic quadrilateral is 180°,

$\implies$ $\angle$ABC + $\angle$ ADC = 180°

$\sf \: \implies \: {60}^{ \circ} \: + \: \angle \: ADC\: = \: {180}^{ \circ}$

$\implies$ $\angle$ADC = 180° - 60°

$\implies \: \boxed{\sf \angle \: ADC \: = \: {120}^{ \circ} }$

Now,

$\because$ AD || BC and DC is the transversal,

$\implies$ $\angle$ADC + $\angle$BCD = 180°

( Sum of co-intertior angles is 180° )

$\implies$ 120° + $\angle$BCD = 180°

$\implies$ $\angle$BCD = 180° - 120°

$\implies$ $\angle$BCD = 60°

$\therefore \: \boxed{ \sf \angle \: BCD \: = \: {60}^{ \circ} }$

So, the correct option is (C) 60°

$\green\star$ Method 2:

Here, we need to extend BC to Point E.

$\because$ We know that the sum of either pair of opposite angles of a cyclic quadrilateral is 180°,

$\implies$ $\angle$ABC + $\angle$ ADC = 180°

$\implies$ 60° + $\angle$ADC = 180°

$\implies$ $\angle$ADC = 180° - 60°

$\implies \: \boxed{\angle \sf \: ADC \: = \: {120}^{ \circ} }$

Now,

$\because$ AD || BC and DC is the transversal,

$\implies$ $\angle$ADC = $\angle$DCE = 120°

( Alternate Interior angles are equal)

Now, $\angle$DCE + $\angle$DCB = 180°

( The sum of angles in a linear pair is 180° )

$\implies$ $\angle$BCD = 180° - $\angle$DCE

$\implies$ $\angle$BCD = 180° - 120°

$\implies \sf \: \boxed{\angle \: \sf BCD \: = \: {60}^{ \circ} }$

$\orange\bigstar$Concepts Used:

• The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.

• Sum of co-intertior angles is 180°

• Alternate Interior Angles are equal.

• The sum of angles in a linear pair is 180°.

$\red\bigstar$More to Know:

• A line which intersects two or more lines at distinct points is called a transversal.

• The corresponding angles are equal.

• Co-intertior angles are also known as consecutive interior angles or allied angles.

• Alternate exterior angles are equal.

• Lines parallel to the same line are parallel to each other