Question

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

Answer 1

∠BPC = 180° - ∠APB              (we are doing this because we know that ∠BPC

                                                 and ∠BPC are supplementary)

180° - ∠APB  = 180° - 100° = 80°

we know that all the triangles made here are isosceles, so that means that ΔBPC is also isosceles Hence,

∠CBP = ∠BCP

let angle CBP be x, So,

∠BCP = x

hence we can write that:-

80° + x° + x° = 180°

⇒ 80° + 2x° = 180°

⇒ 2x° = 180° - 80°             (transposing 80° to RHS)

⇒2x° = 100°

⇒ 2x°/2 = 100°/2               (dividing both the sides by 2)

⇒ x° = 50°

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