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Question

Quadrilateral upper A upper B upper C upper D is drawn within a circle such that all vertices touch the edge of the circle.

The m∠B is 93° , mBC is 58° , and mCD is 106° . Find the missing angle measures.

(3 points)
m∠A=
°, m∠C=
°, and m∠D=
°

Answer 1

In a cyclic quadrilateral (one where all vertices lie on a circle), the sum of the opposite angles is equal to 180°. This means:

1. \( m\angle A + m\angle C = 180^\circ \)
2. \( m\angle B + m\angle D = 180^\circ \)

From the problem, we are given:

• \( m\angle B = 93^\circ \)
• \( m\angle C \) (unknown yet)
• \( mCD = 106^\circ \) (which is the angle opposite angle B)
• \( m\angle D \) (unknown yet)

First, we can use the second equation to find \( m\angle D \):

\[ m\angle B + m\angle D = 180^\circ \implies 93^\circ + m\angle D = 180^\circ \] \[ m\angle D = 180^\circ - 93^\circ = 87^\circ \]

Now we have \( m\angle D = 87^\circ \).

Next, we can use the first equation to find \( m\angle C \):

We also know the sum of angles in triangle ABC is \( m\angle C + mCD + mBC = 180^\circ \)

We have:

\[ m\angle C + 106^\circ + 58^\circ = 180^\circ \] \[ m\angle C + 164^\circ = 180^\circ \] \[ m\angle C = 180^\circ - 164^\circ = 16^\circ \]

Now we can find \( m\angle A \):

From earlier, \( m\angle A + m\angle C = 180^\circ \)

So,

\[ m\angle A + 16^\circ = 180^\circ \] \[ m\angle A = 180^\circ - 16^\circ = 164^\circ \]

Thus, the missing angle measures are:

• \( m\angle A = 164^\circ \)
• \( m\angle C = 16^\circ \)
• \( m\angle D = 87^\circ \)

In summary:

\( m\angle A = 164^\circ \), \( m\angle C = 16^\circ \), and \( m\angle D = 87^\circ \).