The angles of a triangle are x, y and 40^{cir | vedclass

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Question

(A) Given that $x, y$ and $40^{\circ}$ are the angles of a triangle.Since the sum of all angles of a triangle is $180^{\circ}$,we have:$x + y + 40^{\c

Vedclass · Accepted Answer

$85^{\circ}$ and $55^{\circ}$

Vedclass · Answer

(A) Given that $x, y$ and $40^{\circ}$ are the angles of a triangle.Since the sum of all angles of a triangle is $180^{\circ}$,we have:$x + y + 40^{\circ} = 180^{\circ}$$\Rightarrow x + y = 140^{\circ} \quad \dots(i)$Also,the difference between the two angles is given as:$x - y = 30^{\circ} \quad \dots(ii)$Adding equations $(i)$ and $(ii)$:$(x + y) + (x - y) = 140^{\circ} + 30^{\circ}$$2x = 170^{\circ}$$x = 85^{\circ}$Substituting $x = 85^{\circ}$ in equation $(i)$:$85^{\circ} + y = 140^{\circ}$$y = 140^{\circ} - 85^{\circ} = 55^{\circ}$Thus,the values of $x$ and $y$ are $85^{\circ}$ and $55^{\circ}$ respectively.

The angles of a triangle are $x, y$ and $40^{\circ}$. The difference between the two angles $x$ and $y$ is $30^{\circ}$. Find $x$ and $y$.

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