By Team Aakash Byju's | 31st December 2022

In the given figure, ABCD is a rectangle in which APB = 100°. Find the value of x.

The value of x in the given image is 50°.

x

50°

Given that the image ABCD represents a rectangle. It is known that the diagonals of a rectangle are congruent and bisect each other.

AC = BD

=>AP = PC = DP = PB

We know that the sum of all interior angles of a triangle is 180°. Also in ΔAPB, AP = PB and the opposite angles of these sides are equal we can write ∠ABP = ∠PAB

Also, ∠APB = 100° and  ∠APB + ∠ABP + ∠PAB = 180°

=> 100 + ∠ABP + ∠PAB = 180° => ∠ABP + ∠ABP = 80°                                        (∵ ∠ABP = ∠PAB)

=> 2∠ABP = 80°  => ∠ABP = ∠PAB = 40°

Since AC is a straight line.  ∠APC = 180°

Further, ∠APB = 100°  => ∠APB + ∠BPC = 180°

=> 100° + ∠BPC = 180°  => ∠BPC = 80°

Since it is noted that PC = BP, the  opposite angles ∠PBC = ∠BCP.

Also, ∠BPC + ∠PBC = ∠BCP = 180°

=> 80° + ∠PBC + ∠PBC = 180°                                     (∵ ∠PBC = ∠BCP)

=> 2∠PBC = 100° => ∠PBC = ∠BCP = 50° => x = ∠BCP = 50°