The angles of a cyclic quadrilateral ABCD

Solution:

We know that, by property of cyclic quadrilateral,

Sum of opposite angles = 180°

∠A + ∠C = (6x + 10)° + (x + y)° = 180°

$\left[\because \angle A=(6 x+10)^{\circ}, \angle C=(x+y)^{\circ}\right.$, given $]$

$\Rightarrow \quad 7 x+y=170$ ...(i)

and $\angle B+\angle D=(5 x)^{\circ}+(3 y-10)^{\circ}=180^{\circ}$

$\left[\because \angle B=(5 x)^{\circ}, \angle D=(3 y-10)^{\circ}\right.$, given $]$

$\Rightarrow$ $5 x+3 y=190^{\circ}$ .....(ii)

On multiplying Eq. (i) by 3 and then subtracting, we get

$3 \times(7 x+y)-(5 x+3 y)=510^{\circ}-190^{\circ}$

$\Rightarrow$ $21 x+3 v-5 x-3 v=320^{\circ}$

$\Rightarrow$ $16 x=320^{\circ}$

$\therefore$ $x=20^{\circ}$

On putting $x=20^{\circ}$ in Eq. (i), we get

$7 \times 20+y=170^{\circ}$

$\Rightarrow$ $y=170^{\circ}-140^{\circ} \Rightarrow y=30^{\circ}$

$\therefore$ $\angle A=(6 x+10)^{\circ}=6 \times 20^{\circ}+10^{\circ}$

$=120^{\circ}+10^{\circ}=130^{\circ}$

$\angle B=(5 x)^{\circ}=5 \times 20^{\circ}=100^{\circ}$

$\angle C=(x+y)^{\circ}=20^{\circ}+30^{\circ}=50^{\circ}$

 

$\angle D=(3 y-10)^{\circ}=3 \times 30^{\circ}-10^{\circ}$

$=90^{\circ}-10^{\circ}=80^{\circ}$

Hence, the required values of x and y are 20° and 30° respectively and the values of the four angles ;.e., ZA, ZB, ZC and ZD are 130°, 100°, 50°

and 80°, respectively.