Question
ABCD is a cyclic quadrilateral such that $\angle\text{A}=(4\text{y}+20)^\circ,\angle\text{B}=(3\text{y}-5)^\circ$ $\angle\text{C}=(-4\text{x})^\circ$ and $\angle\text{D}=(7\text{x}+5)^\circ$ Find the four angles.
✓
Answer
We know that the sum of the opposite angles of cyclic quadrilateral is 180° in the cyclic quadrilateral ABCD angle A and C angles B and D pairs of opposite angles
$\therefore\angle\text{A}+\angle\text{C}=180^\circ$ and $\angle\text{B}+\angle\text{D}=180^\circ$
$\angle\text{A}+\angle\text{C}=180^\circ$
By Substituting $\angle\text{A}=(4\text{y}+20)^\circ$ and $\angle\text{C}=(-4\text{x})^\circ$ we get
4y + 20° - 4x = 180°
-4x + 4y + 20° = 180°
-4x + 4y = 180° - 20°
-4x + 4y = 160°
4x - 4y = -160°
Divide both sides of equation by 4 we get
x - y = -40°
x - y + 40° = 0 .....(i)
$\angle\text{B}+\angle\text{D}=180^\circ$
BY substituting $\angle\text{B}=(3\text{y}-5)^\circ$ and $\angle\text{D}=(7\text{x}+5)^\circ$ we get
3y - 5° + 7x + 5° = 180°
7x + 3y = 180°
7x + 3y - 180° = 0 ......(ii)
By multiplying equation (i) by 3 we get
3x - 3y + 120° = 0 .....(iii)
By adding equation (iii) from (ii) we get
$3\text{x}\ -\ 3\text{y}\ +\ 120\ =\ 0\\7\text{x}\ +\ 3\text{y}\ -\ 180\ =\ 0\over10\text{x}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 60$
$\text{x}=\frac{60}{10}$
$\text{x}=6^\circ$
By substituting x = 6° in equation (i) we get
x - y + 40° = 0
6° - y + 40° = 0
-1y = -40° - 6°
-1y = -46°
$\text{y}=\frac{-46^\circ}{-1}$
$\text{y}=46^\circ$
The angles of a cyclic quadrilateral are
$\angle\text{A}=4\text{y}+20$
= (4 × 46)° + 20°
= 184 + 20
= 204°
$\angle\text{B}=3\text{y}-5$
= 3 × 46 - 5
= 138 - 5 = 133°
$\angle\text{C}=-4\text{x}^\circ$
= -[4(6)]°
= -24°
$\angle\text{D}=\big(7\text{x}+5\big)^\circ$
= (7 × 6)° + 5°
= (42 + 5)°
= 47°
Hence the angles of quadrilateral are $\angle\text{A}=204^\circ,\angle\text{B}=133^\circ\angle\text{C}=-24^\circ,\angle\text{D}=47^\circ$
$\therefore\angle\text{A}+\angle\text{C}=180^\circ$ and $\angle\text{B}+\angle\text{D}=180^\circ$
$\angle\text{A}+\angle\text{C}=180^\circ$
By Substituting $\angle\text{A}=(4\text{y}+20)^\circ$ and $\angle\text{C}=(-4\text{x})^\circ$ we get
4y + 20° - 4x = 180°
-4x + 4y + 20° = 180°
-4x + 4y = 180° - 20°
-4x + 4y = 160°
4x - 4y = -160°
Divide both sides of equation by 4 we get
x - y = -40°
x - y + 40° = 0 .....(i)
$\angle\text{B}+\angle\text{D}=180^\circ$
BY substituting $\angle\text{B}=(3\text{y}-5)^\circ$ and $\angle\text{D}=(7\text{x}+5)^\circ$ we get
3y - 5° + 7x + 5° = 180°
7x + 3y = 180°
7x + 3y - 180° = 0 ......(ii)
By multiplying equation (i) by 3 we get
3x - 3y + 120° = 0 .....(iii)
By adding equation (iii) from (ii) we get
$3\text{x}\ -\ 3\text{y}\ +\ 120\ =\ 0\\7\text{x}\ +\ 3\text{y}\ -\ 180\ =\ 0\over10\text{x}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 60$
$\text{x}=\frac{60}{10}$
$\text{x}=6^\circ$
By substituting x = 6° in equation (i) we get
x - y + 40° = 0
6° - y + 40° = 0
-1y = -40° - 6°
-1y = -46°
$\text{y}=\frac{-46^\circ}{-1}$
$\text{y}=46^\circ$
The angles of a cyclic quadrilateral are
$\angle\text{A}=4\text{y}+20$
= (4 × 46)° + 20°
= 184 + 20
= 204°
$\angle\text{B}=3\text{y}-5$
= 3 × 46 - 5
= 138 - 5 = 133°
$\angle\text{C}=-4\text{x}^\circ$
= -[4(6)]°
= -24°
$\angle\text{D}=\big(7\text{x}+5\big)^\circ$
= (7 × 6)° + 5°
= (42 + 5)°
= 47°
Hence the angles of quadrilateral are $\angle\text{A}=204^\circ,\angle\text{B}=133^\circ\angle\text{C}=-24^\circ,\angle\text{D}=47^\circ$
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