Question
In a cyclic quadrilateral ABCD, it is given that $\angle\text{A}=(\text{2x}+4)^\circ,\ \angle\text{B}=(\text{y}+3)^\circ,$ $\angle\text{C}=(\text{2y}+10)^\circ$ and $\angle\text{D}=(\text{4x}-5)^\circ$ Find the four angles.
✓
Answer
Given that in a cyclic quadrilateral ABCD,
$\angle\text{A}=(\text{2x}+4)^\circ,\ \angle\text{B}=(\text{y}+3)^\circ,$ $\angle\text{C}=(\text{2y}+10)^\circ$ and $\angle\text{D}=(\text{4x}-5)^\circ$
We know that,
Opposite angles of a quadrilateral sum upto 180°
$\Rightarrow\angle\text{B}+\angle\text{D}=180^\circ$
⇒ (y + 3)° + (4x - 5)° = 180°
⇒ 4x + y = 182 ...(i)
Similarly, $\angle\text{A}+\angle\text{C}=180^\circ$
⇒ (2x + 4)° + (2y + 10)° = 180°
⇒ 2x + 2y = 166
⇒ x + y = 83 ...(i)
Subtracting (ii) from (i), we get
⇒ 3x = 99
⇒ x = 33
Substituting x = 33 in (ii), we get
⇒ y = 50
Hence, the angles of ABCD are
So, $\angle\text{A}=70^\circ,\ \angle\text{B}=53^\circ,$ $\angle\text{C}=110^\circ$ and $\angle\text{D}=127^\circ$
$\angle\text{A}=(\text{2x}+4)^\circ,\ \angle\text{B}=(\text{y}+3)^\circ,$ $\angle\text{C}=(\text{2y}+10)^\circ$ and $\angle\text{D}=(\text{4x}-5)^\circ$
We know that,
Opposite angles of a quadrilateral sum upto 180°
$\Rightarrow\angle\text{B}+\angle\text{D}=180^\circ$
⇒ (y + 3)° + (4x - 5)° = 180°
⇒ 4x + y = 182 ...(i)
Similarly, $\angle\text{A}+\angle\text{C}=180^\circ$
⇒ (2x + 4)° + (2y + 10)° = 180°
⇒ 2x + 2y = 166
⇒ x + y = 83 ...(i)
Subtracting (ii) from (i), we get
⇒ 3x = 99
⇒ x = 33
Substituting x = 33 in (ii), we get
⇒ y = 50
Hence, the angles of ABCD are
So, $\angle\text{A}=70^\circ,\ \angle\text{B}=53^\circ,$ $\angle\text{C}=110^\circ$ and $\angle\text{D}=127^\circ$
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