Find the arcs (O is the center of the circle).

Question

Вопрос:

Find the arcs (O is the center of the circle).

Ответ:

ФИО:Телефон:Емаил:Полное описание сути нарушения прав (почему распространение данной информации запрещено Правообладателем):

Accepted Answer

Solution:

In each case, we need to find the measure of an arc. The measure of a central angle is equal to the measure of its intercepted arc. The measure of an inscribed angle is half the measure of its intercepted arc. The measure of an arc subtended by a chord is related to the angle subtended by the chord at the center.

Problem	Diagram Description	Arc Calculation	Answer
1	Central angle \(\angle BOC = 120^\)	Arc BC = \(\angle BOC\)	\(120^\)
2	Inscribed angle \(\angle BAC = 40^\) intercepts arc BC.	Arc BC = \(2 \cdot \angle BAC\)	\(2 \cdot 40^ = 80^\)
3	Central angle \(\angle AOC = x\) and \(\angle ABC = x\) intercept arc AC. This implies it is an isosceles triangle AOC. Angle ABC is inscribed.	Arc AC = \(\angle AOC\) and Arc AC = \(2 \cdot \angle ABC\). So \(\angle AOC = 2x\). In \(\triangle AOC\), \(\angle OAC = \angle OCA = (180 - 2x)/2 = 90 - x\). We need more information to solve this. Assuming \(x\) is the angle subtended by arc AC at the circumference.	Cannot be determined with the given information.
4	Central angle subtended by arc CD is \(\angle COD = x\). Inscribed angle \(\angle CAD = x\). Right angle at A, \(\angle DAB = 90^\) and \(\angle BAC = 40^\).	Arc CD = \(\angle COD\) and Arc CD = \(2 \cdot \angle CAD\). If \(\angle BAC = 40^\), then arc BC = \(2 \cdot 40^ = 80^\). Since \(\angle DAB = 90^\), arc DB = \(2 \cdot 90^ = 180^\). But \(\angle DAB\) is not a central or inscribed angle for arc DB. If \(\angle BCD = 90^\) then BD is diameter. Arc BD = \(180^\). If \(\angle ADB = 90^\) then AB is diameter. If \(\angle CDA = 90^\) then AC is diameter. Assuming \(x\) is the inscribed angle subtended by arc CD. So arc CD = \(2x\). Given \(\angle BAC = 40^\), arc BC = \(2 \times 40^ = 80^\).	Cannot be determined with the given information.
5	Inscribed angle \(\angle BXC = 110^\). O is the center. This angle is not an inscribed angle related to a simple arc. If \(\angle BOC = 110^\) then arc BC = \(110^\). If \(\angle BAC = 110^\) that would be incorrect as an inscribed angle cannot be obtuse. If \(\angle BDC = 110^\) also incorrect. If \(\angle ABC = 110^\) also incorrect. It is likely that \(\angle BOC = 110^\).	Arc BC = \(\angle BOC\)	\(110^\)
6	Central angle \(\angle AOC = 100^\). X is a point on the arc AC. \(\angle ABC\) is the inscribed angle subtending arc AC.	Arc AC = \(\angle AOC\)	\(100^\)
7	Inscribed angles \(\angle CAD = x\) and \(\angle BAC = 30^\).	Arc CD = \(2 \cdot \angle CAD = 2x\). Arc BC = \(2 \cdot \angle BAC = 2 \times 30^ = 60^\).	Cannot be determined with the given information.
8	Inscribed angle \(\angle BAC = 30^\). O is the center. \(\angle BOC\) is the central angle subtending arc BC.	Arc BC = \(2 \times \angle BAC = 2 \times 30^ = 60^\). Arc BC = \(\angle BOC\).	\(60^\)
9	Inscribed angle \(\angle BAC = x\). \(\angle ABC = 35^\). Arc BC = \(2x\). Arc AC = \(2 \times 35^ = 70^\).	Arc BC = \(2x\).	Cannot be determined with the given information.
10	\(\angle BAE = 25^\). Point E is on the arc BC. Arcs BE and EC are marked with tick marks, implying they are equal. \(\angle AOE\) and \(\angle EOC\) are likely related to the equal arcs.	Arc BE = Arc EC. \(\angle BAE = 25^\) subtends arc BE. So arc BE = \(2 \times 25^ = 50^\). Since arc BE = arc EC, arc EC = \(50^\). Arc BC = arc BE + arc EC = \(50^ + 50^ = 100^\). Also \(\angle BCE = \angle BAE = 25^\). And \(\angle CBE = \angle CAE\). \(\angle BAC\) subtends arc BC, so \(\angle BAC = 100^ / 2 = 50^\). \(\angle ABC\) subtends arc AC.	Arc BC = \(100^\).
11	Inscribed angles \(\angle BAC = x\) and \(\angle CAD = x\). \(\angle CBD = 40^\).	Arc BC = \(2x\). Arc CD = \(2 \times 40^ = 80^\). Arc BD = Arc BC + Arc CD = \(2x + 80^\). \(\angle BAD = 2x\). \(\angle BCD\) subtends arc BD.	Cannot be determined with the given information.
12	\(\angle BAC = x\). \(\angle BCA = 35^\). \(\angle CAD = 30^\). O is the center.	Arc BC = \(2x\). Arc AB = \(2 \times 35^ = 70^\). Arc CD = \(2 \times 30^ = 60^\). Arc BD = Arc BC + Arc CD = \(2x + 60^\). Arc AC = Arc AB + Arc BC = \(70^ + 2x\). Also \(\angle BOC = 2x\). \(\angle AOC = 2 \times \angle ABC\). \(\angle AOB = 2 \times \angle ACB = 2 \times 35^ = 70^\). This matches arc AB. \(\angle COA = 2 \times \angle CBA\). \(\angle COD = 2 \times \angle CAD = 2 \times 30^ = 60^\). This matches arc CD. The sum of arcs is \(360^\). Arc AB + Arc BC + Arc CD + Arc DA = \(360^\). \(70^ + 2x + 60^ + \text{Arc DA} = 360^\). \(130^ + 2x + \text{Arc DA} = 360^\).	Arc BC = \(2x\).

Note: Some problems could not be solved with the given information.