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Home » 9. In a cyclic quadrilateral ABCD, ∠A = (2x + 4)o, ∠B = (y + 3)o, ∠C = (2y + 10)o, ∠D = (4x – 5)o. Find the four angles.
9. In a cyclic quadrilateral ABCD, ∠A = (2x + 4)o, ∠B = (y + 3)o, ∠C = (2y + 10)o, ∠D = (4x – 5)o. Find the four angles.
CBSE | Class 10 | Exercise 3.11 | Maths | Pair of Linear Equations In Two Variables | RD Sharma
9. In a cyclic quadrilateral ABCD, ∠A = (2x + 4)o, ∠B = (y + 3)o, ∠C = (2y + 10)o, ∠D = (4x – 5)o. Find the four angles.

Solution:

The sum of the opposite angles of cyclic quadrilateral should be 180o.

And, in the cyclic quadrilateral ABCD,

Angles

    \[\angle A\]

and

    \[\angle C\]

& angles

    \[\angle B\]

and

    \[\angle D\]

are the pairs of opposite angles.

    \[\angle A\text{ }+\angle C\text{ }=\text{ }{{180}^{o}}\]

    \[\angle B\text{ }+\angle D\text{ }=\text{ }{{180}^{o}}\]

Substituting the value of two equations

For

    \[\angle A\text{ }+\angle C\text{ }=\text{ }{{180}^{o}}\]

    \[\Rightarrow \angle A\text{ }=\text{ }{{\left( 2x\text{ }+\text{ }4 \right)}^{o}}~\]

and

    \[\angle C\text{ }=\text{ }{{\left( 2y\text{ }+\text{ }10 \right)}^{o}}\]

    \[2x\text{ }+\text{ }4\text{ }+\text{ }2y\text{ }+\text{ }10\text{ }=\text{ }{{180}^{o}}\]

    \[2x\text{ }+\text{ }2y\text{ }+\text{ }14\text{ }=\text{ }{{180}^{o}}\]

    \[2x\text{ }+\text{ }2y\text{ }=\text{ }{{180}^{o~}}\text{ }{{14}^{o}}\]

    \[2x\text{ }+\text{ }2y\text{ }=\text{ }166\text{ }.............\left( i \right)\]

And,

    \[\angle B\text{ }+\angle D\text{ }=\text{ }{{180}^{o}}\]

,

⇒

    \[\angle B\text{ }=\text{ }{{\left( y+3 \right)}^{o}}~\]

and

    \[\angle D\text{ }=\text{ }{{\left( 4x\text{ }\text{ }5 \right)}^{o}}\]

    \[y\text{ }+\text{ }3\text{ }+\text{ }4x\text{ }\text{ }5\text{ }=\text{ }{{180}^{o}}\]

    \[4x\text{ }+\text{ }y\text{ }\text{ }5\text{ }+\text{ }3\text{ }=\text{ }{{180}^{o}}\]

    \[4x\text{ }+\text{ }y\text{ }\text{ }2\text{ }=\text{ }{{180}^{o}}\]

    \[4x\text{ }+\text{ }y\text{ }=\text{ }{{180}^{o}}~+\text{ }{{2}^{o}}\]

    \[4x\text{ }+\text{ }y\text{ }=\text{ }{{182}^{o}}~...........\left( ii \right)\]

Solve (i) & (ii)

Multiplying equation (ii) by 2,

    \[8x\text{ }+\text{ }2y\text{ }=\text{ }364\text{ }...........(iii)\]

Subtract equation (iii) from (i)

    \[-6x\text{ }=\text{ }-198\]

    \[x\text{ }=\text{ }-198/\text{ }-6\]

    \[\Rightarrow x\text{ }=\text{ }{{33}^{o}}\]

Substituting the value of

    \[x\text{ }=\text{ }{{33}^{o}}\]

in equation (ii) to find y

    \[4x\text{ }+\text{ }y\text{ }=\text{ }182\]

    \[132\text{ }+\text{ }y\text{ }=\text{ }182\]

    \[y\text{ }=\text{ }182\text{ }\text{ }132\]

    \[\Rightarrow y\text{ }=\text{ }50\]

Calculating the angles of a cyclic quadrilateral :

    \[\angle A\text{ }=\text{ }2x\text{ }+\text{ }4\]

    \[=\text{ }66\text{ }+\text{ }4\]

    \[=\text{ }{{70}^{o}}\]

    \[\angle B\text{ }=\text{ }y\text{ }+\text{ }3\]

    \[=\text{ }50\text{ }+\text{ }3\]

    \[=\text{ }{{53}^{o}}\]

    \[\angle C\text{ }=\text{ }2y\text{ }+\text{ }10\]

    \[=\text{ }100\text{ }+\text{ }10\]

    \[=\text{ }{{110}^{o}}\]

    \[\angle D\text{ }=\text{ }4x\text{ }\text{ }5\]

    \[=\text{ }132\text{ }\text{ }5\]

    \[=\text{ }{{127}^{o}}\]

The angles of the cyclic quadrilateral ABCD are

    \[\angle A\text{ }=\text{ }{{70}^{o}},\]

    \[\angle B\text{ }=\text{ }{{53}^{o}},\]

    \[\angle C\text{ }=\text{ }{{110}^{o}}\]

and

    \[\angle D\text{ }=\text{ }{{127}^{o}}\]

i

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