Question

In the adjoining figure, ◻ABCD is a square and ΔABP is an equilateral triangle.

Find :

1. ∠AOB
2. ∠CPB
3. ∠PCD

Answer 1

Given:

ABCD is a square.

ΔABP is equilateral.

O is the intersection of the diagonals center of the square.

Place coordinates:

A(0, 1)

B(0, 0) 

C(1, 0)

D(1, 1) 

And Side = 1

Then `P = (sqrt(3)/2, 1/2)`

Step-wise calculation:

1. ∠AOB (Vertex at O):

`O = (1/2, 1/2)`

OA = A – O

= `(-1/2, 1/2)`

OB = B – O

= `(-1/2, -1/2)`

Dot product OA·OB

= `(-1/2)(-1/2) + (1/2)(-1/2)`

= 0

Zero dot product

⇒ OA ⟂ OB

⇒ ∠AOB = 90°

2. ∠CPB (Vertex at P):

`P = (h, 1/2)` with `h = sqrt(3)/2 ≈ 0.8660`.

PB = B – P

= `(-h, -1/2)`

PC = C − P

= `(1 - h, -1/2)`

`PB·PC = (-h)(1 - h) + (-1/2)(-1/2)`

= `-h(1 - h) + 1/4`

With `h = sqrt(3)/2`.

Evaluate: PB·PC = 0.133975...

|PB| = 1   ...`("Since"  PB^2 = h^2 + (1/2)^2 = 3/4 + 1/4 = 1)` 

`|PC| = sqrt( (1 - h)^2 + (1/2)^2) ≈ 0.5177`

`cos ∠CPB = (PB·PC)/(|PB||PC|)`

= `0.133975/0.5177`

= 0.258819

⇒ ∠CPB = arccos(0.258819)

⇒ ∠CPB = 75°

3. ∠PCD (Vertex at C):

CD vector = D – C

= (0, 1)

CP = P – C

= `(h - 1, 1/2)`

= `(-0.133975, 1/2)`

Dot CD·CP

= `0 xx (h - 1) + 1 xx (1/2)`

= `1/2`

|CP| ≈ 0.5177

|CD| = 1 

⇒ cos ∠PCD = `(1/2)/0.5177 ≈ 0.9659258`

⇒ ∠PCD ≈ 15°

These 75° and 15° values match standard worked examples for a square with an equilateral triangle on one side.

1. ∠AOB = 90°
2. ∠CPB = 75°
3. ∠PCD = 15°