Answer:
Step-by-step explanation:
The central angle is double the inscribed angle for the same intercepted arc. Since doubling the angle adds 30° to it, the original inscribed angle must be 30°.
Then the central angle is 30°+30° = 2·30° = 60°.
You can use the definition of inscribed angle along with the theorem that angle inscribed by an arc on the circumference is half of what angle it subtends on the center.
The measure of angles are:
[tex]\angle AOB = 60^\circ\\[/tex]
Measure of inscribed angle by the arc AB is 30°
Inscribed angle made by the arc is the angle made by two chords in that circle whose one end lies on endpoints of that arc and another end is joined and lies on the circumference.
Let AB represents an arc in a circle with center at O. Then there are two cuts on the circle at A and B. There are two arcs AB, one is minor and one is major.
If the angle inscribed by AB lies on major arc, then it is half of the angle subtended by that arc on the center of the circle.
If the angle inscribed by AB lies on minor arc, then it is supplement of the half of the angle subtended by that arc on the center of the circle.
Supplement of an angle = 180 degrees - that angle.
For given condition, let the point on circumference where the inscribed angle is made is called P.
Let the angle subtended by arc AB is denoted by x°
Case 1: P is on minor arc:
Since angle AOB is x°, and since it is 30 degrees greater than the inscribed angle, thus, we have angle APB = x° - 30 (so that angle AOB stays 30° greater than the inscribed angle APB)
Then we have
[tex]\angle APB = 180^\circ - \dfrac{\angle AOB}{2}\\x - 30 = 180 - \dfrac{x}{2}\\\\2x - 60 = 360 -x\\3x = 420\\\\x = \dfrac{420}{3} = 140^\circ[/tex]
Thus, [tex]\angle AOB = 140^\circ[/tex] and [tex]\angle APB = x^\circ - 30^\circ = 140 - 30 = 110^\circ[/tex]
Case 2: P is on major arc:
Since angle AOB is x°, and since it is 30 degrees greater than the inscribed angle, thus, we have angle APB = x° - 30 (so that angle AOB stays 30° greater than the inscribed angle APB)
[tex]\angle APB = \dfrac{\angle AOB}{2}\\x - 30 = \dfrac{x}{2}\\\\2x - 60 = x\\x = 60[/tex]
Thus, we have
[tex]\angle AOB = x^\circ = 60^\circ[/tex] and [tex]\angle APB = x^\circ - 30^\circ = 60^\circ - 30^\circ = 30^\circ[/tex]
Both the angles are plotted on below given graph.
Learn more about inscribed angle theorem here:
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