12 multiple-choice questions, progressively harder.
Which description defines an inscribed angle?
Solution
Correct answer: B
An inscribed angle needs a vertex on the circle and two chord rays through two other distinct points. The other choices fail the vertex or chord conditions.
Points A,V,B,WA,V,B,WA,V,B,W occur in that order around a circle. Which arc is intercepted by ∠AVB\angle AVB∠AVB?
Correct answer: D
The intercepted arc must join A to B without passing through V. Only arc AWB has those endpoints and excludes V.
An inscribed angle intercepts an arc of 132∘132^\circ132∘. What is its measure?
Correct answer: A
Halve the intercepted arc.
An inscribed angle measures 41∘41^\circ41∘. What is the measure of its intercepted arc?
Correct answer: C
Double the inscribed angle.
The major arc AXBAXBAXB excluding VVV measures 238∘238^\circ238∘. Find m∠AVBm\angle AVBm∠AVB.
Use the stated arc, which excludes the vertex.
The angle measures 119 degrees.
ABABAB is a diameter, and CCC is another point on the circle. Find m∠ACBm\angle ACBm∠ACB.
The intercepted arc is a semicircle.
The angle is 90 degrees.
Distinct points A,B,CA,B,CA,B,C lie on a circle. If m∠ABC=90∘m\angle ABC=90^\circm∠ABC=90∘, which segment is a diameter?
Angle ABC intercepts a semicircle with endpoints A and C. The semicircle converse makes AC a diameter.
P,QP,QP,Q lie on the same open arc between A,BA,BA,B. If m∠APB=57∘m\angle APB=57^\circm∠APB=57∘, find m∠AQBm\angle AQBm∠AQB.
Both vertices exclude the same arc AB, so their angles are equal.
P,RP,RP,R lie on opposite open arcs between A,BA,BA,B. If m∠APB=39∘m\angle APB=39^\circm∠APB=39∘, find m∠ARBm\angle ARBm∠ARB.
Their intercepted arcs total a full circle, so their angles total 180 degrees.
The angle is 141 degrees.
Minor arc TATATA measures 116∘116^\circ116∘. Ray TRTRTR is tangent at TTT on the same side of line OTOTOT as AAA. Find m∠RTAm\angle RTAm∠RTA.
The named tangent ray selects the minor arc.
The angle is 58 degrees.
Major arc TBATBATBA measures 226∘226^\circ226∘. Ray TSTSTS is tangent at TTT on the opposite side of line OTOTOT from AAA. Find m∠STAm\angle STAm∠STA.
The opposite tangent ray selects the major arc.
The angle measures 113 degrees.
VAVAVA is a diameter with center OOO. Another point BBB lies on the circle, and m∠VBO=32∘m\angle VBO=32^\circm∠VBO=32∘. Find m∠AVBm\angle AVBm∠AVB.
Radii OV and OB make triangle VOB isosceles, so its angle at V is also 32 degrees. Ray VA is ray VO, making this the requested angle.
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