Question

Find the four angles of a cyclic quadrilateral ABCD in which \[\angle A = {(2x - 1)^0},\angle B = {(y + 5)^0},\angle C = {(2y + 15)^0},\angle D = {(4x - 7)^0}\].

Answer 1

Hint: Here we will find the value of x and y using the property of cyclic quadrilateral i.e., the sum of opposite angles is equal to ${180^0}$.

Complete step-by-step answer:
As you know that, in a cyclic quadrilateral sum of opposite angles is equal to ${180^0}$.
So,
 $
  \angle A + \angle C = {180^0} \\
   \Rightarrow 2x - 1 + 2y + 15 = {180^0} \\
   \Rightarrow x + y = {83^0} \to (1) \\
$
Similarly
$
  \angle B + \angle D = {180^0} \\
   \Rightarrow y + 5 + 4x - 7 = {180^0} \\
   \Rightarrow 4x + y = {182^0} \to (2) \\
$
Solving equation (1) and equation (2) we get
$
 \Rightarrow 3x = {99^0} \\
  \Rightarrow x = {33^0} \\
 $
Putting x in equation (1) we get $y = {50^0}$
So angles are:
$
  \angle A = {65^0} \\
  \angle B = {55^0} \\
  \angle C = {115^0} \\
  \angle D = {125^0} \\
$

Note: To do these types of questions it is important to remember that a cyclic quadrilateral the sum of opposite angles is equal to ${180^0}$. Then equating the equations yields the answer.