Question

1. a) triangle abc is inscribed in circle g. find the measures of angles a, b and c. b) triangle que is inscribed in circle o. if the measure of arc eq is 144 degrees, find the measure of angle u.

Explanation:

Part A: Triangle \( ABC \) inscribed in circle \( G \)

Step 1: Identify the diameter and semicircle

Since \( AC \) passes through the center \( G \), \( AC \) is the diameter of the circle. A diameter divides the circle into two semicircles, each measuring \( 180^\circ \). The arc \( AB \) is \( 120^\circ \), arc \( BC \) is \( 60^\circ \), and arc \( AC \) (the semicircle) would be \( 180^\circ \), which checks out (\( 120^\circ + 60^\circ = 180^\circ \)).

Step 2: Use the inscribed angle theorem

The inscribed angle theorem states that an angle inscribed in a semicircle is a right angle. So, \( \angle ABC \) is inscribed in the semicircle \( AC \), so \( \angle ABC = 90^\circ \).

Step 3: Find \( \angle BAC \) (angle \( A \))

The measure of an inscribed angle is half the measure of its intercepted arc. For \( \angle BAC \) (angle \( A \)), it intercepts arc \( BC \) which is \( 60^\circ \). So, \( \angle BAC=\frac{1}{2}\times60^\circ = 30^\circ \).

Step 4: Find \( \angle BCA \) (angle \( C \))

For \( \angle BCA \) (angle \( C \)), it intercepts arc \( AB \) which is \( 120^\circ \). So, \( \angle BCA=\frac{1}{2}\times120^\circ = 60^\circ \).

Step 5: Verify using triangle angle sum

The sum of angles in a triangle is \( 180^\circ \). \( \angle A + \angle B + \angle C = 30^\circ + 90^\circ + 60^\circ = 180^\circ \), which is correct.

Step 1: Find the measure of arc \( QU \) or arc \( QE \)? Wait, first, the total circumference of a circle is \( 360^\circ \), but we know arc \( EQ = 144^\circ \). Wait, actually, triangle \( QUE \) is inscribed in the circle, and we need to find \( \angle U \). Let's recall the inscribed angle theorem again.

Wait, maybe we need to find the intercepted arc. Wait, in triangle \( QUE \), angle \( U \) intercepts arc \( QE \)? Wait, no. Wait, the inscribed angle theorem: the measure of an inscribed angle is half the measure of its intercepted arc.

Wait, let's assume that \( QE \) is an arc, and we need to find angle \( U \). Wait, maybe the triangle is isoceles? Wait, no, let's think again. Wait, the circle is \( O \), and triangle \( QUE \) is inscribed. Wait, maybe the arc opposite to angle \( U \) is arc \( QE \). Wait, the measure of angle \( U \) is half the measure of its intercepted arc. Wait, if arc \( EQ = 144^\circ \), then the intercepted arc for angle \( U \) is arc \( EQ \)? Wait, no, angle \( U \) is at point \( U \), so the intercepted arc is arc \( QE \). Wait, the inscribed angle theorem says that the measure of an inscribed angle is half the measure of its intercepted arc. So, \( \angle U=\frac{1}{2}\times\text{measure of arc } QE \). Wait, but arc \( QE = 144^\circ \), so \( \angle U=\frac{1}{2}\times144^\circ = 72^\circ \)? Wait, no, wait, maybe the arc is the major arc or the minor arc? Wait, if arc \( EQ \) is \( 144^\circ \), then the minor arc \( EQ \) is \( 144^\circ \), so the inscribed angle over it would be half. Wait, let's confirm.

Wait, in circle \( O \), triangle \( QUE \) is inscribed. So, angle \( U \) is an inscribed angle intercepting arc \( QE \). So, the measure of angle \( U \) is half the measure of arc \( QE \). So, \( \angle U=\frac{1}{2}\times144^\circ = 72^\circ \).

Answer:

• \( \angle A = 30^\circ \)
• \( \angle B = 90^\circ \)
• \( \angle C = 60^\circ \)

Part B: Triangle \( QUE \) inscribed in circle \( O \)