Questions & Answers

Question

Answers

Answer
Verified

Given data:

$\angle BPC = {124^o}$................. (1)

Now as we see from the figure that APC is the diagonal of the rectangle which is a straight line so it makes a straight angle with the point p.

Now as we know that the straight angle = 180 degrees.

So, $\angle BPC + \angle BPA = {180^o}$

Now from equation (1) we have,

$ \Rightarrow {124^o} + \angle BPA = {180^o}$

$ \Rightarrow \angle BPA = {180^o} - {124^o} = {56^o}$............ (2)

Now as we know that in a rectangle the length of the diagonals are always equal and also bisect each other.

Therefore, PA = PB = PC = PB.

So triangles ABP, BCP, CDP, and DAP are an isosceles triangle.

Now as we know that in a isosceles triangle angles opposite to equal sides are always equal.

So in triangle ABP, as AP = BP, so $\angle PBA = \angle BAP$.............. (3)

Now as we know that in a triangle the sum of all angles are equal to 180 degrees.

So in triangle ABP we have,

$ \Rightarrow \angle PBA + \angle BAP + \angle BPA = {180^o}$

Now from equation (2) and (3) we have,

$ \Rightarrow \angle BAP + \angle BAP + {56^o} = {180^o}$

$ \Rightarrow \angle BAP = \dfrac{{{{180}^o} - {{56}^o}}}{2}$

$ \Rightarrow \angle BAP = \dfrac{{{{124}^o}}}{2} = {62^o}$

$ \Rightarrow \angle PBA = \angle BAP = {62^o}$

Now as we know in a rectangle opposite sides are equal and parallel to each other.

Therefore, $\angle PBA = \angle PDC = {62^o}$............. (4) (alternate angles)

Now as we know that in a rectangle adjacent sides of the rectangle always makes a 90 degrees.

Therefore, $\angle ADC = {90^o}$

$ \Rightarrow \angle ADP + \angle PDC = {90^o}$

Now from equation (4) we have,

$ \Rightarrow \angle ADP + {62^o} = {90^o}$

$ \Rightarrow \angle ADP = {90^o} - {62^o} = {28^o}$

×

Sorry!, This page is not available for now to bookmark.