Geometry Problem: Rotation of Triangle in Circle and Arc Measures

Published on May 20, 2026

Geometry

Solve a geometry problem involving the rotation of a triangle inscribed in a circle, using arc measures to find an unknown angle. Demonstrates properties of rotations and inscribed angles.

Question

Answer

Based on the problem description and the provided image, here is the step-by-step solution:

1. Analyze the given information

is inscribed in a circle.
The arc measures are given as: and .
The total degrees in a circle is , so the measure of is:However, for the angle , we use the minor arc . The inscribed angle is half the measure of its intercepted arc :

2. Understand the rotation

is rotated clockwise around point to become .
Point also lies on the same circle.
Since and are on the circle, and and are on the circle, and (because it’s a rotation), the chord is equal to the chord .
In a circle, equal chords subtend equal arcs. Therefore, .
Let the rotation angle be . This is the angle .

3. Determine the rotation angle

Since are on the circle and , the triangle is an isosceles triangle with .
The angle is the angle of rotation.
From the figure, is the point on the circle such that the chord is the same length as .
The angle is an inscribed angle that intercepts .
Wait, a simpler way: Since , the arcs they subtend from the center are equal. The angle is an inscribed angle.
Actually, let’s look at the angle . The problem states is the angle between and .
Since is rotated to , the angle .
The rotation angle is .
From the diagram, the angle is .
is not correct based on the visual. Looking at the shaded region, is the angle between the original side and the new side .
is not correct either.
Let’s re-examine: (if is “inside” the angle ).
Actually, the rotation angle . Since , the arc equals arc .
The angle is an inscribed angle. The arc is the arc between and .
Since , the points and are symmetric with respect to the diameter passing through .
The angle is the angle between the chord and the chord .
In a circle, the angle between two equal chords sharing a vertex is (if we consider the center). More simply, .
Since , the arc (minor) is the same as arc (minor).
Let the minor arc . Then is the major arc, so the minor arc .
The angle is an inscribed angle. The arc is .
Thus, . This is the rotation angle .

4. Calculate

The angle is .
(as is between and in the rotation).
.
.

Comparing this to the options:
(A) 27
(B) 31
© 32
(D) 37

The value of

is 37.

Final Answer: (D) 37

Theorems

• Inscribed Angle Theorem

An inscribed angle is half of the central angle that subtends the same arc. Or, the measure of an inscribed angle is half the measure of its intercepted arc. $ ext{Inscribed Angle} = rac{1}{2} imes ext{Intercepted Arc}$

• Arc Measure of Full Circle

The sum of the measures of the arcs in a circle is

. $ ext{arc } AB + ext{arc } BC + ext{arc } CA = 360^ ext{o}$

• Rotation Properties

Rotation is a rigid transformation that preserves distances and angles. When a figure is rotated, corresponding sides and angles remain congruent. If $ riangle ABC$ is rotated to $ riangle AB’C’$, then

, and $ ext{arc } AB = ext{arc } AB’$, $ ext{arc } AC = ext{arc } AC’$, $ ext{arc } BC = ext{arc } B’C’$. The angle of rotation is $ heta = ext{arc } BB’ = ext{arc } CC’$. Also, $ ext{arc } AB = ext{arc } AB’$ implies that $ riangle ABB’$ is isosceles. The angle $ heta$ is the angle of rotation, e.g. $ heta = ext{arc } BB’$ where the arc is subtended by an angle at the center, or $ heta = rac{1}{2} ext{arc } BB’$ if considering inscribed angle.

• Isosceles Triangle Properties

In an isosceles triangle, the sides opposite the equal angles are equal, and the angles opposite the equal sides are equal. If

, then $ riangle ABB’$ is isosceles, and the base angles $ riangle BAB’$ are equal. The angle $ heta$ is the angle of rotation.

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