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ABCD is cyclic quadrilateral with $∠ ADC = 135^{\circ}$ , sides BA and CD produced meet at point P, sides AD and BC produced meet at point Q. If $∠ P : ∠ Q = 2 : 1$ , find angles P and Q.

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30^{\circ}

and

45^{\circ}

45^{\circ}

and

60^{\circ}

60^{\circ}

and

30^{\circ}

90^{\circ}

and

60^{\circ}

answer is C.

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Detailed Solution

It is given that,

P : Q = 2 : 1

Suppose the ratio constant = x

So,

∠P = 2x and ∠Q = x
Also, in the cyclic quadrilateral ABCD, we have
∠ADC

= 135^{\circ}

From the cyclic quadrilateral property

∠ B = 180^{\circ} - ∠ ADC

\Rightarrow ∠ B = 180^{\circ} - 135^{\circ}

\Rightarrow ∠ B = 45^{\circ}

Since, in a linear pair, sum of angles

= 180^{\circ}

∠ ADC + ∠ ADP = 180^{\circ}

\Rightarrow ∠ ADP = 180^{\circ} - ∠ ADC

\Rightarrow ∠ ADP = 180^{\circ} - 135^{\circ}

\Rightarrow ∠ ADP = 45^{\circ}

Now, in ΔAQB from angle sum property of triangle

\Rightarrow ∠ A + ∠ Q + ∠ B = 180^{\circ}

\Rightarrow ∠ A + x + 45^{\circ} = 180^{\circ}

\Rightarrow ∠ A = 180^{\circ} - (x + 45^{\circ})

\Rightarrow ∠ A = (135^{\circ} - x) \dots \dots \dots . . (1)

In ΔADP,
We know that,
A triangle's exterior angle = product of its diagonal interior angles.
Exterior angle

A = ∠ ADP + ∠ P

∠ A = 45^{\circ} + 2 x \dots \dots \dots (2)

Equate the value of angle A from equation (1) and 2

\Rightarrow 45^{\circ} + 2 x = 135^{\circ} - x

(Taking terms of x to one side)

\Rightarrow 3 x = 135^{\circ} - 45^{\circ}

\Rightarrow 3 x = 90^{\circ}

\Rightarrow x = 30^{\circ}

So the angles,

∠ Q = 30^{\circ}

∠ P = 2 x = 60^{\circ}

Hence, option 3 is correct.

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