Since quadrilateral PQRS is inscribed in circle O, the opposite angles P and R must add up to 180 degrees (as per the inscribed angle theorem).
We have the measures of angles P and R given as:
- Angle P = (x + 16)°
- Angle R = (6x - 4)°
Setting up the equation based on the property of opposite angles:
\[ (x + 16) + (6x - 4) = 180 \]
Simplifying this equation:
\[ x + 16 + 6x - 4 = 180 \] \[ 7x + 12 = 180 \] \[ 7x = 180 - 12 \] \[ 7x = 168 \] \[ x = \frac{168}{7} = 24 \]
Now, substituting \( x = 24 \) back into the expressions for angles P and R:
\[ \text{Angle P} = x + 16 = 24 + 16 = 40° \] \[ \text{Angle R} = 6x - 4 = 6(24) - 4 = 144 - 4 = 140° \]
Now, we need to determine the measures of the other two angles, Q and S. Since angles P and R are known, we can find:
\[ \text{Angle Q} + \text{Angle S} = 180° \]
And since opposite angles in a cyclic quadrilateral equal 180°, we have:
- Angle Q = Angle R = 140°
- Angle S = Angle P = 40°
We've calculated Angle P = 40° and Angle R = 140°.
The smaller angle among the four angles (P, Q, R, S) is:
\[ \text{Smaller Angle} = \text{Angle P} = 40° \]
Thus, the measure of the smaller angle is 40°.
Final answer: 40°.
1 answer