Question - Cyclic Quadrilateral Angle Calculation
Solution:
In a cyclic quadrilateral, the sum of opposite angles is 180 degrees.
Given angle K = 120 degrees, we find the opposite angle M, which is 110 degrees + x.
120 degrees + (110 degrees + x) = 180 degrees
230 degrees + x = 180 degrees
x = 180 degrees - 230 degrees
x = -50 degrees
Since an angle cannot be negative in this context, we need to reconsider the equation for opposite angles.
We use the property that the exterior angle is equal to the opposite interior angle.
Thus the exterior angle at vertex L is 120 degrees which is equal to angle N + angle M.
Angle N is 70 degrees because it is the base angle of the isosceles triangle KLN (angle K is 120 degrees).
120 degrees = 70 degrees + angle M
Angle M = 120 degrees - 70 degrees
Angle M = 50 degrees
Therefore, x = 50 degrees
Tutor Solution:
In a cyclic quadrilateral, the opposite angles sum up to \( 180^\circ \). This means that in quadrilateral \( KLMN \), we have:
\[ \angle K + \angle M = 180^\circ \] \[ \angle L + \angle N = 180^\circ \]We are given that \( \angle K = 120^\circ \) and \( \angle M = 110^\circ \), thus:
\[ 120^\circ + 110^\circ = 230^\circ \]This does not leave us with a direct answer because these are not the opposite angles. We need to use another property of the cyclic quadrilateral, which is that the exterior angle is equal to the opposite interior angle. Here, \( \angle L \) is the exterior angle to \( \angle N \).
Since \( \angle N + x = 110^\circ \) (because \( x \) is the exterior angle to \( \angle N \)), we can say:
\[ \angle N = 110^\circ - x \]Now, let’s calculate \( \angle L \), knowing the sum of \( \angle L \) and \( \angle N \):
\[ \angle L + \angle N = 180^\circ \] \[ \angle L + 110^\circ - x = 180^\circ \] \[ \angle L = 180^\circ - 110^\circ + x \] \[ \angle L = 70^\circ + x \]But since \( \angle L \) is the same as the sum of \( \angle K \) and \( \angle M \), we use the sum of \( \angle K \) and \( \angle M \) to find \( x \):
\[ 70^\circ + x = 120^\circ + 110^\circ - 180^\circ \] \[ 70^\circ + x = 230^\circ - 180^\circ \] \[ 70^\circ + x = 50^\circ \] \[ x = 50^\circ - 70^\circ \] \[ x = -20^\circ \]Since a negative angle doesn't make sense in this context, we have made a mistake. Let's re-evaluate our approach. The correct way is to set up the equation \( \angle L + \angle N = 180^\circ \) directly, knowing that both of them are interior angles of the cyclic quadrilateral:
\[ \angle L = 70^\circ + x \] \[ \angle N = 110^\circ \]Now we solve for \( x \) using:
\[ 70^\circ + x + 110^\circ = 180^\circ \] \[ x + 180^\circ = 180^\circ \] \[ x = 180^\circ - 180^\circ \] \[ x = 0^\circ \]This is clearly not one of our answer choices and again suggests a mistake. We should instead use the property of the cyclic quadrilateral correctly:
\[ \angle L + 110^\circ = 180^\circ \] \[ \angle L = 180^\circ - 110^\circ \] \[ \angle L = 70^\circ \]Now remembering that \( \angle L \) is the exterior angle to \( \angle KMN \), and \( x \) is the interior opposite angle, they must be equal:
\[ x = \angle L \]Therefore, the value of \( x \) is:
\[ x = 70^\circ \]The correct answer is:
\[ \boxed{x = 70^\circ} \]So the choice (A) 70° is the correct answer.
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