12 multiple-choice questions, progressively harder.
Triangle PQRPQRPQR has QRQRQR extended beyond RRR to SSS, with Q,R,SQ,R,SQ,R,S in that order. Which are the remote interior angles for ∠PRS\angle PRS∠PRS?
Solution
Correct answer: C
The exterior angle has vertex RRR. Its remote interior angles are the triangle's interior angles at the other vertices, PPP and QQQ.
The interior angle ∠PRQ\angle PRQ∠PRQ is adjacent to the exterior angle, so it is not remote.
Triangle ABCABCABC has BCBCBC extended beyond CCC to DDD, with B,C,DB,C,DB,C,D in that order. If m∠BAC=26∘m\angle BAC=26^\circm∠BAC=26∘ and m∠ABC=83∘m\angle ABC=83^\circm∠ABC=83∘, what is m∠ACDm\angle ACDm∠ACD?
Correct answer: A
The angles at AAA and BBB are remote interior angles for the exterior angle at CCC. Add their measures by the exterior angle theorem.
26∘+83∘=109∘26^\circ+83^\circ=109^\circ26∘+83∘=109∘
Triangle ABCABCABC has BCBCBC extended beyond CCC to DDD, with B,C,DB,C,DB,C,D in that order. If m∠ACD=142∘m\angle ACD=142^\circm∠ACD=142∘ and m∠BAC=68∘m\angle BAC=68^\circm∠BAC=68∘, what is m∠ABCm\angle ABCm∠ABC?
Correct answer: D
The exterior measure is the sum of the two remote measures.
142∘=68∘+m∠ABC142^\circ=68^\circ+m\angle ABC142∘=68∘+m∠ABC
Subtract the known remote measure to get m∠ABC=74∘m\angle ABC=74^\circm∠ABC=74∘. The 38∘38^\circ38∘ supplement of the exterior angle is instead the adjacent angle at CCC.
Triangle ABCABCABC has BCBCBC extended beyond CCC to DDD, with B,C,DB,C,DB,C,D in that order. If m∠ACB=113∘m\angle ACB=113^\circm∠ACB=113∘, what is m∠ACDm\angle ACDm∠ACD?
Correct answer: B
The given interior angle is adjacent to the exterior angle. Their noncommon rays CBCBCB and CDCDCD are opposite, so they form a linear pair.
m∠ACD=180∘−113∘=67∘m\angle ACD=180^\circ-113^\circ=67^\circm∠ACD=180∘−113∘=67∘
A triangle has a side extended at one vertex. The remote interior angles for that exterior angle measure 44∘44^\circ44∘ and 63∘63^\circ63∘. What is its measure?
The given angles are the two remote interior angles. Their sum is the exterior measure, not its supplement.
44∘+63∘=107∘44^\circ+63^\circ=107^\circ44∘+63∘=107∘
A triangle's exterior angle has measure eee, and its remote interior angles have measures aaa and bbb. Which must be true?
The exterior angle theorem gives e=a+be=a+be=a+b. Both remote measures are positive in a triangle.
Subtract each remote measure in turn.
e−a=b>0∘e-a=b>0^\circe−a=b>0∘
e−b=a>0∘e-b=a>0^\circe−b=a>0∘
Both differences are positive, proving both inequalities.
A path follows the three sides of a triangle once, taking one non-reflex left exterior turn at every vertex. Two turns measure 151∘151^\circ151∘ and 104∘104^\circ104∘. What is the third?
These are three consistently oriented exterior turns, one per vertex, so their measures total 360∘360^\circ360∘.
360∘−151∘−104∘=105∘360^\circ-151^\circ-104^\circ=105^\circ360∘−151∘−104∘=105∘
Triangle ABCABCABC has BCBCBC extended beyond CCC to DDD, with B,C,DB,C,DB,C,D in that order. If m∠ACD=123∘m\angle ACD=123^\circm∠ACD=123∘, what is m∠ACBm\angle ACBm∠ACB?
The adjacent interior and exterior angles form a linear pair because B,C,DB,C,DB,C,D occur in that order on a line.
m∠ACB=180∘−123∘=57∘m\angle ACB=180^\circ-123^\circ=57^\circm∠ACB=180∘−123∘=57∘
At one vertex of a triangle, either incident side can be extended beyond the vertex, making two non-reflex exterior angles. How are their measures related?
Let the interior measure be ccc. Each side extension makes its exterior angle a linear pair with that same interior angle.
Each exterior measure is the same supplement.
e1=180∘−ce_1=180^\circ-ce1=180∘−c
e2=180∘−ce_2=180^\circ-ce2=180∘−c
Thus the two exterior angles have equal measures. The choices can represent opposite turning directions.
In triangle ABCABCABC, a distinct point DDD lies outside the triangle. Which extra condition makes ∠ACD\angle ACD∠ACD the exterior angle formed by continuing side BCBCBC past CCC?
A side continuation must lie on the original side's line, beyond its endpoint. The order B,C,DB,C,DB,C,D makes rays CBCBCB and CDCDCD opposite.
That order gives the linear pair between ∠ACB\angle ACB∠ACB and ∠ACD\angle ACD∠ACD. Merely placing DDD outside the triangle does not give it.
Triangle ABCABCABC has BCBCBC extended beyond CCC to DDD, with B,C,DB,C,DB,C,D in that order. The measures in degrees are m∠BAC=2x+7m\angle BAC=2x+7m∠BAC=2x+7, m∠ABC=3x+8m\angle ABC=3x+8m∠ABC=3x+8, and m∠ACD=7x−5m\angle ACD=7x-5m∠ACD=7x−5. Which equation follows directly from the exterior angle theorem?
The angles at AAA and BBB are remote; their sum equals the exterior angle at CCC.
(2x+7)+(3x+8)=7x−5(2x+7)+(3x+8)=7x-5(2x+7)+(3x+8)=7x−5
The exterior angle is not one of the triangle's three interior angles. Neither remote angle is its adjacent linear-pair partner.
An exterior angle of a triangle measures 90∘90^\circ90∘. One of its remote interior angles measures 36∘36^\circ36∘. What is the other remote measure?
The remote measures add to the exterior measure.
90∘−36∘=54∘90^\circ-36^\circ=54^\circ90∘−36∘=54∘
The adjacent interior angle is also 90∘90^\circ90∘, but it is not the requested remote angle.
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