In thegiven figure, ABCD is a rectangle in which ∠APB = 100°. The value of x is
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∠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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So the answer for this question is
x = c) 50 degrees
x = c) 50 degrees
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