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# The angles of a triangle are x, y, and ${40^o}$. The difference between the two angles x and y is ${30^o}$. Find x and y.$A.\;\;\;\;\;55^\circ ,85^\circ \\ B.\;\;\;\;\;22^\circ ,67^\circ \\ C.\;\;\;\;\;87^\circ ,29^\circ \\ D.\;\;\;\;\;66^\circ ,76^\circ \\$

Last updated date: 13th Jun 2024
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Hint: As we have,${\text{Sum of all angles of a triangle = 18}}{{\text{0}}^{\text{o}}}$ using this approach, we will substitute the values of angles of a triangle to get an equation in x and y and we have an equation given as the difference of x and y is given, then we solve them by substitution method to get the required result.

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, option (A) is correct.

Note: It was assumed that $x > y$, but it is not mentioned in the question so after getting the values of x and y, we can’t say the which one will be greater as if we assume that $x < y$
$y - x = {30^o} \\ y = {30^o} + x......(iii) \\$
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 ${85^o}$
Additional information: When it is provided that the difference between any two quantity, we cannot just say which one will be greater or lesser, so we use the expression of modulus$\left( {|x|} \right).$