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ABCD is a cyclic quadrilateral such that $\angle A = {\left( {4y + 20} \right)^\circ },\angle B = {\left( {3y - 5} \right)^\circ },\angle C = 4{x^\circ }$ and$\angle D = {\left( {7x + 5} \right)^\circ }$. Find the four angles.

Answer
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Hint: Cyclic Quadrilateral is a quadrilateral whose all arms on verities lie on a single circle. Key concept to be applied here is that the sum of four angles is ${360^\circ }$.Also sum of opposite angles in a Cyclic Quadrilateral is ${180^\circ }$.

Complete step-by-step answer:
Let us assume ABCD is the cyclic quadrilateral, $\angle A = {\left( {4y + 20} \right)^\circ },\angle B = {\left( {3y - 5} \right)^\circ },\angle C = 4{x^\circ },\angle D = {\left( {7x + 5} \right)^\circ }$
We know that the addition of four angle of a quadrilateral = ${360^\circ }$
$
  \therefore \angle A + \angle B + \angle C + \angle D = {360^\circ } \\
   \Rightarrow 4y + {20^\circ } + 3y - {5^\circ } + 4{x^\circ } + 7x + {5^\circ } = {360^\circ } \\
   \Rightarrow 7y + 11x = {360^\circ } - {20^\circ } \\
   \Rightarrow 7y + 11x = {340^\circ }.............\left( I \right) \\
 $
Also, the sum of opposite angles in a Cyclic Quadrilateral is ${180^\circ }$.
$\angle A + \angle C = {180^\circ }$ or $\angle B + \angle D = {180^\circ }$
$
  \therefore 4y + 20 + 4x = {180^\circ } \\
   \Rightarrow 4x + 4y = {160^\circ }............\left( {II} \right) \\
 $
Now,
$
  11x + 7y = 340.........(I) \times 4 \\
  4x + 4y = 160..........\left( {II} \right) \times 7 \\
  44x + 28y = 1360 \\
  28x + 28y = 1120 \\
   \Rightarrow 16x = 240 \\
   \Rightarrow x = \dfrac{{240}}{{16}} \\
   \Rightarrow x = 15 \\
 $
Putting the value of x in the equation II, we will get the value of y as follows:
$
   \Rightarrow 4x + 4y = 160 \\
   \Rightarrow 4 \times 15 + 4y = 160 \\
   \Rightarrow 60 + 4y = 160 \\
   \Rightarrow 4y = 160 - 60 = 100 \\
   \Rightarrow y = \dfrac{{100}}{4} = 25 \\
 $
Now we will substitute the values of x and y to get all four angles as follows:
$
  \angle A = 4y + 20 = 4 \times 25 + 20 = {120^\circ } \\
  \angle B = 3y - 5 = 3 \times 25 - 5 = {70^\circ } \\
  \angle C = 4x = 4 \times 15 = {60^\circ } \\
  \angle D = 7x + 5 = 7 \times 15 + 5 = {110^\circ } \\
 $

So, four angles of ABCD are ${120^\circ }, {70^\circ }, {60^\circ } and {110^\circ }.$

Note: An important property of a cyclic quadrilateral is that the sum of opposite angles is always equal to ${180^\circ }$. This property is used in the above problem. This property is useful in geometrical proofs and properties.