Question

quadrilateral PQRS is inscribed in circle O. opposite angles P and R measure (x + 16) and (6x - 4), respectively, Determine the measure of the smaller angle

Answer 1

Since quadrilateral PQRS is inscribed in circle O, opposite angles are supplementary. This means that the sum of angles P and R is \(180^\circ\).

Given the angles:
- Angle P: \(x + 16\)
- Angle R: \(6x - 4\)

We set up the equation based on the supplementary angles:

\[
(x + 16) + (6x - 4) = 180
\]

Now, combine like terms:

\[
x + 16 + 6x - 4 = 180
\]

\[
7x + 12 = 180
\]

Next, subtract 12 from both sides:

\[
7x = 180 - 12
\]

\[
7x = 168
\]

Now, divide by 7:

\[
x = \frac{168}{7}
\]

\[
x = 24
\]

Now that we have the value of \(x\), we can find the measures of angles P and R:

1. Calculate angle P:

\[
\text{Angle P} = x + 16 = 24 + 16 = 40^\circ
\]

2. Calculate angle R:

\[
\text{Angle R} = 6x - 4 = 6(24) - 4 = 144 - 4 = 140^\circ
\]

Now we find the smaller angle between angle P and angle R.

The angles are:
- Angle P: \(40^\circ\)
- Angle R: \(140^\circ\)

Thus, the smaller angle is:

\[
\boxed{40^\circ}
\]