In a cyclic quadrilateral ABCD, angle A = (2x + 4)°, angle B = (y + 3)°, angle C = (2y + 10)° and angle D = (4x – 5)°. Find the smallest angle.

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angle A

angle B

angle C

angle D

answer is B.

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Detailed Solution

As we know, the sum of the opposite angles of a cyclic quadrilateral is always 180°.

https://www.vedantu.com/question-sets/8b39b358-1140-474a-9856-27f3a2f408fb7769242660460932115.png

Therefore, angle A + angle C = 180°
And, angle B + angle D = 180°
According to the question,
Angle A = (2x + 4)°
Angle B = (y + 3)°
Angle C = (2y + 10)°
Angle D = (4x – 5)°
Therefore, (2x + 4) + (2y + 10) = 180°
And (y + 3) + (4x – 5) = 180°
Let us solve (2x + 4) + (2y + 10) = 180° first.
⇒ 2x + 4 + 2y + 10 = 180
⇒ 2x + 2y + 14 = 180
⇒ 2x + 2y = 166
⇒ 2x = 166 – 2y
x =

\frac{166 - 2 y}{2}

⇒ y =

\frac{166 - 2 x}{2}

Let us solve (y + 3) + (4x – 5) = 180° now.
⇒ y + 3 + 4x – 5 = 180
⇒ y + 4x – 2 = 180
⇒ y + 4x = 182
⇒ 4x = 182 – y
x =

\frac{182 - y}{4}

⇒ y = 182 – 4x
Let us now equate the values of ‘x’.

\frac{166 - 2 y}{2}

\frac{182 - y}{4}

⇒ 4 (166 – 2y) = 2 (182 – y)
⇒ 664 – 8y = 364 – 2y
⇒ 664 – 364 = -2y + 8y
⇒ 300 = 6y
⇒ 50 = y
Let us now equate the values of ‘y’.

\frac{166 - 2 y}{2}

= 182 – 4x
⇒ 166 – 2x = 2 (182 – 4x)
⇒ 166 – 2x = 364 – 8x
⇒ -2x + 8x = 364 – 166
⇒ 6x = 198
⇒ x = 33
Now, let us substitute the values of ‘x’ and ‘y’ in the measures of the angles of the quadrilateral ABCD.
Angle A = (2x + 4)° = (2 × 33 + 4) = (66 + 4) = 70°
Angle B = (y + 3)° = (50 + 3) = 53°
Angle C = (2y + 10)° = (2 × 50 + 10) = (100 + 10) = 110°
Angle D = (4x – 5)° = (4 × 33 - 5) = (132 – 5) = 127°
So, the smallest angle, i.e. the angle with the smallest measure is Angle B, whose measure is 53°.
Option (2) is correct.

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