Question

The angles of a cyclic quadrilateral ABCD are $$A = \left( {6x + 10} \right),B = \left( {5x} \right),C = \left( {x + y} \right),D = \left( {3y - 10} \right)$$.
Find x and y, and hence the values of the four angles.
(A) $$x = 10,y = 30,A = {120^ \circ },B = {100^ \circ },C = {50^ \circ },D = {60^ \circ }$$
(B) $$x = 20,y = 30,A = {130^ \circ },B = {100^ \circ },C = {50^ \circ },D = {80^ \circ }$$
(C) $$x = 18,y = 30,A = {140^ \circ },B = {100^ \circ },C = {50^ \circ },D = {84^ \circ }$$
(D) $$x = 30,y = 30,A = {180^ \circ },B = {100^ \circ },C = {50^ \circ },D = {80^ \circ }$$

Answer 1

Hint: We are given a cyclic quadrilateral, therefore we plot the given points and form a quadrilateral. To find the values of x and y, we use the opposite angles rule in a quadrilateral. After we get the values of x and y, we can substitute in the angles A, B, C and D to the angles.

Complete step-by-step solution:
Let us consider the given points and draw a quadrilateral,
$$A = \left( {6x + 10} \right),B = \left( {5x} \right),C = \left( {x + y} \right),D = \left( {3y - 10} \right)$$

We know that, in a quadrilateral, the sum of the opposite angles of the quadrilateral is equal to $${180^ \circ }$$
So, in the above given quadrilateral, we can use this rule to find out the value of x
$$ \Rightarrow \angle A + \angle C = {180^ \circ }$$
Now, let us substitute the values given,
$$ \Rightarrow 6x + {10^ \circ } + x + y = {180^ \circ }$$
$$ \Rightarrow 7x + y = {170^ \circ }$$……. (1)
Now, there are other two opposite angles,
$$ \Rightarrow \angle B + \angle D = {180^ \circ }$$
Substituting the given angles,
$$ \Rightarrow 3y - {10^ \circ } + 5x = {180^ \circ }$$
$$ \Rightarrow 5x + 3y = {190^ \circ }$$……. (2)
Now, let us solve the equations (1) and (2)
We multiply equation (1) by $$3$$
$$\eqalign{
  & 7x + y = {170^ \circ } \times 3 \cr
  &\Rightarrow 5x + 3y = {190^ \circ } \cr} $$
We get,
$$\eqalign{
  & 21x + 3y = {510^ \circ } \cr
  &\Rightarrow 5{x_{\left( - \right)}} + 3{y_{\left( - \right)}} = {190^ \circ }_{\left( - \right)} \cr} $$
Now, we simplify the above equation
$$16x = {320^ \circ }$$
Simplifying the equation, we get the value of x,
$$ \Rightarrow x = {20^ \circ }$$
From equation (1),
$$y = {170^ \circ } - 7x$$
Now that we know the value of x, we can substitute it to find the value of y.
$$ \Rightarrow y = {170^ \circ } - 7\left( {{{20}^ \circ }} \right)$$
$$ \Rightarrow y = {170^ \circ } - {140^ \circ }$$
$$ \Rightarrow y = {30^ \circ }$$
Therefore, $$x = {20^ \circ },y = {30^ \circ }$$
Let us substitute the value of x and y to get the angles,
Given that,
$$A = \left( {6x + 10} \right)$$
$$ \Rightarrow A = 6\left( {{{20}^ \circ }} \right) + {10^ \circ }$$
$$ \Rightarrow A = {130^ \circ }$$
Now let us find out the angle B
$$B = \left( {5x} \right)$$
$$ \Rightarrow B = 5\left( {{{20}^ \circ }} \right)$$
$$ \Rightarrow B = {100^ \circ }$$
For angle C,
$$C = \left( {x + y} \right)$$
$$ \Rightarrow C = \left( {{{20}^ \circ } + {{30}^ \circ }} \right)$$
$$ \Rightarrow C = {50^ \circ }$$
For angle D, we have
$$D = \left( {3y - 10} \right)$$
$$ \Rightarrow D = 3\left( {{{30}^ \circ }} \right) - {10^ \circ }$$
$$ \Rightarrow D = {80^ \circ }$$
Now we have the final answer,
$$x = 20,y = 30,A = {130^ \circ },B = {100^ \circ },C = {50^ \circ },D = {80^ \circ }$$
Hence, option (B) is correct.

Note: Learn the opposite angles rule for the quadrilateral since it is the main theorem applied in the question. Always draw a rough diagram so that you get an idea of the problem. There are two parts in the question, so do not stop after finding x and y, find the values of A, B, C and D as well.