ABCD is a cyclic quadrilateral (see figure). Find the angles of the cyclic quadrilateral.
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 13"

AcademicMathematicsNCERTClass 10


ABCD is a cyclic quadrilateral.

To do:

We have to find the angles of the cyclic quadrilateral.


We know that,

The sum of the opposite angles of a cyclic quadrilateral is $180^{\circ}$.

This implies,

$\angle \mathrm{B}+\angle \mathrm{D} =180^{\circ}$

$3 y-5^{\circ}+(-7 x+5^{\circ}) =180^{\circ}$

$3 y-5^{\circ}-7 x+5^{\circ} =180^{\circ}$

$-7 x+3 y =180^{\circ}$

$7 x-3 y =-180^{\circ}$...........(i)


$\angle \mathrm{A}+\angle \mathrm{C}=180^{\circ}$

$4 y+20^{\circ}+(-4x) =180^{\circ}$


$4 x-4 y =-160^{\circ}$

$4(x-y) =4(-40^{\circ})$


Multiplying (ii) by 3 and subtracting the result from (i), we get,

$7x-4 x=-180^{\circ}+120^{\circ}$

$4 x=-60^{\circ}$



Substituting of $x=-15^{\circ}$ in (ii), we get,




Hence, the angles of the cyclic quadrilateral are:

$\angle \mathbf{A}=4 y+20^{\circ}$

$=4 \times 25^{\circ}+20^{\circ}$


$\angle \mathbf{B}=3 y-5^{\circ}$

$=3 \times 25^{\circ}-5^{\circ}$


$\angle \mathbf{C}=-4 x$

$=-4 \times-15^{\circ}$


$\angle \mathbf{D}=-7 x+5^{\circ}$

$=-7 \times-15^{\circ}+5^{\circ}$


Updated on 10-Oct-2022 13:20:08