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Pair of Linear Equations in Two variables — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsPair of Linear Equations in Two variables5 Marks
Question
The angles of a cyclic quadrilateral ABCD are
$\angle\text{A}=(6\text{x}+10)^\circ,\ \angle\text{B}=(5\text{x})^\circ$
$ \angle\text{C}=(\text{x}+\text{y})^\circ, \angle\text{D}=(3\text{y}-10)^\circ$
Find x and y, and hence the values of the four angles.
✓

Answer

We know that, by property of cuclic quadrilateral,Sum of opposite angles = 180°
$\big[\because\ \angle\text{A}=(6\text{x}+10)^\circ,\ \angle\text{C}=(\text{x}+\text{y})^\circ,\ \text{given}\big]$ $\Rightarrow\ 7\text{x}+\text{y}=170\ .....(\text{i})$ and $\angle\text{B}+\angle\text{D}=(5\text{x})^\circ+(3\text{y}-10)^\circ=180^\circ$ $\big[\because\ \angle\text{B}=(5\text{x})^\circ,\ \angle\text{D}=(3\text{y}-10)^\circ,\ \text{given}\big]$ $\Rightarrow\ 5\text{x}+3\text{y}=190^\circ\ .....(\text{ii})$ On multiplying Eq. (i) by 3 and then subtracting, we get $3\times(7\text{x}+\text{y})-(5\text{x}+3\text{y})=510^\circ-190^\circ$ $\Rightarrow\ 21\text{x}+3\text{y}-5\text{x}-3\text{y}=320^\circ$ $\Rightarrow\ 16\text{x}=320^\circ$ $\therefore\ \text{x}=20^\circ$ On putting $\text{x}=20^\circ$ in Eq. (i), we get $7\times20+\text{y}=170^\circ$ $\Rightarrow\ \text{y}=170^\circ-140^\circ\Rightarrow\ \text{y}=30^\circ$ $\therefore\ \angle\text{A}=(6\text{x}+10)^\circ=6\times20^\circ+10^\circ$ $=120^\circ+10^\circ=130^\circ$ $\angle\text{B}=(5\text{x})^\circ=5\times20^\circ=100^\circ$ $\angle\text{C}=(\text{x}+\text{y})^\circ=20^\circ+30^\circ=50^\circ$ $\angle\text{D}=(3\text{y}-10)^\circ=3\times30^\circ-10^\circ$ $=90^\circ-10^\circ=80^\circ$ Hence, the required values of x and y are 20° and 30° respectively and the values of the four angles i.e., ZA, ZB, ZC, and ZD are 130°, 100°, 50°and 80° respectively.

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