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Linear Equations In Two Variables — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsLinear Equations In Two Variables5 Marks
Question
Find the angles of a cyclic quadrilateral ABCD in which:
$\angle\text{A}=(\text{4x}+20)^\circ,$ $\angle\text{B}=(\text{3x}-5)^\circ,$ $\angle\text{C}=(\text{4y})^\circ$ and $\angle\text{D}=(\text{7y}+5)^\circ.$
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Answer

Given:
In a cyclic quadrilateral ABCD, we have:
$\angle\text{A}=(4\text{x}+20)^\circ$
$\angle\text{B}=(3\text{x}-5)^\circ$
$\angle\text{C}=(4\text{y})^\circ$
$\angle\text{D}=(7\text{y}+5)^\circ$
$\angle\text{A}+\angle\text{C}=180^\circ$ and $\angle\text{B}+\angle\text{D}=180^\circ$ [Since ABCD is a cyclic quadrilateral]
Now, $\angle\text{A}+\angle\text{C}=(4\text{x}+20)^\circ+(4\text{y})^\circ=180^\circ$
⇒ 4x + 4y + 20 = 180
⇒ 4x + 4y = 180 - 20 = 160
⇒ x + y = 40 ....(i)
Also, $\angle\text{B}+\angle\text{D}=(3\text{x}-5)^\circ+(7\text{y}+5)^\circ=180^\circ$
⇒ 3x + 7y = 180 ...(ii)
On multiplying (i) by 3, we get:
⇒ 3x + 3y = 120 ...(iii)
On subtracting (iii) from (ii), we get:
4y = 60
⇒ y = 15
On subtracting y = 15 in (1), we get:
x + 15 = 40
⇒ x = (40 - 15) = 25
Therefore, we have:
$\angle\text{A}=(4\text{x}+20)^\circ=(4\times25+20)^\circ=120^\circ$
$\angle\text{B}=(3\text{x}-5)^\circ=(3\times25-5)^\circ=70^\circ$
$\angle\text{C}=(4\text{y})^\circ=(4\times15)^\circ=60^\circ$
$\angle\text{D}=(7\text{y}+5)^\circ=(7\times15+5)^\circ=(105+5)^\circ=110^\circ$

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