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Linear Equation In Two Variable — Maths STD 10 — Question

Gujarat BoardEnglish MediumSTD 10MathsLinear Equation In Two Variable4 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, $\angle\text{A}+\angle\text{C}=(4\text{x}+20)^\circ+(4\text{y})^\circ=180^\circ$
$\Rightarrow 4x + 4y + 20 = 180$
$\Rightarrow 4x + 4y = 180 - 20 = 160$
$\Rightarrow x + y = 40 ....(i)$
Also, $\angle\text{B}+\angle\text{D}=(3\text{x}-5)^\circ+(7\text{y}+5)^\circ=180^\circ$
$\Rightarrow 3x + 7y = 180 ...(ii)$
On multiplying $(i)$ by $3$, we get:
$\Rightarrow 3x + 3y = 120 ...(iii)$
On subtracting $(iii)$ from $(ii)$, we get:
$4y = 60$
$\Rightarrow y = 15$
On subtracting $y = 15$ in $(1),$ we get:
$x + 15 = 40$
$\Rightarrow x = (40 - 15) = 25$
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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