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\[

A.\;\;\;\;\;55^\circ ,85^\circ \\

B.\;\;\;\;\;22^\circ ,67^\circ \\

C.\;\;\;\;\;87^\circ ,29^\circ \\

D.\;\;\;\;\;66^\circ ,76^\circ \\

\]

Answer

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It is given that, $|x - y| = 30$

Let us assume that $x > y$

Therefore,

$

x - y = {30^o} \\

x = {30^o} + y......(i) \\

$

Since we know that sum of all the angles of a triangle is 180°, we can as

$x + y + {40^o} = {180^o}.......(ii)$

From equation (i) and (ii), we get

\[

{30^o} + y + y + {40^o} = {180^o} \\

\Rightarrow 2y = {180^o} - {40^o} - {30^o} \\

\Rightarrow 2y = {110^o} \\

\Rightarrow y = {55^o} \\

\]

Substituting the value of y in equation (i), we get

$

x = {30^o} + {55^o} \\

\Rightarrow x = {85^o} \\

$

Therefore, angles x and y are \[{55^o}\] and \[{85^o}\]

Hence,

$

y - x = {30^o} \\

y = {30^o} + x......(iii) \\

$

Since we know that sum of all the angles of a triangle is 180°, we can as

$x + y + {40^o} = {180^o}$

From equation (ii) and (iii), we get

\[

{30^o} + x + x + {40^o} = {180^o} \\

\Rightarrow 2x = {180^o} - {40^o} - {30^o} \\

\Rightarrow 2x = {110^o} \\

\Rightarrow x = {55^o} \\

\]

Substituting the value of x in equation (i), we get

$

y = {30^o} + {55^o} \\

\Rightarrow y = {85^o} \\

$

Therefore, angles x and y are \[{55^o}\] and \[{85^o}\]