12 multiple-choice questions, progressively harder.
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=5x−2m\angle BAC=5x-2m∠BAC=5x−2, m∠ABC=2x+9m\angle ABC=2x+9m∠ABC=2x+9, and m∠ACD=10x−17m\angle ACD=10x-17m∠ACD=10x−17. What is xxx?
Solution
Correct answer: B
The remote sum equals the exterior measure.
(5x−2)+(2x+9)=10x−17(5x-2)+(2x+9)=10x-17(5x−2)+(2x+9)=10x−17
Combining terms gives 7x+7=10x−177x+7=10x-177x+7=10x−17, so 24=3x24=3x24=3x and x=8x=8x=8. The remote measures are 38∘38^\circ38∘ and 25∘25^\circ25∘, giving exterior 63∘63^\circ63∘ and adjacent interior 117∘117^\circ117∘; all are valid.
The two remote interior angles for a triangle's 119∘119^\circ119∘ exterior angle have measures in the ratio 2:52:52:5. What are those two measures?
Correct answer: D
The remote measures total 119∘119^\circ119∘. Write them as 2k2k2k and 5k5k5k, with kkk positive.
7k=119∘7k=119^\circ7k=119∘
Thus k=17∘k=17^\circk=17∘, giving 34∘34^\circ34∘ and 85∘85^\circ85∘. Their ratio is 2:52:52:5, and their sum is 119∘119^\circ119∘.
One non-reflex exterior angle at vertex AAA of triangle ABCABCABC measures 148∘148^\circ148∘, and one at BBB measures 107∘107^\circ107∘. What is a non-reflex exterior angle measure at CCC?
Correct answer: A
Each exterior angle supplements its adjacent interior angle. The interior measures at AAA and BBB are 32∘32^\circ32∘ and 73∘73^\circ73∘.
m∠ACB=180∘−32∘−73∘=75∘m\angle ACB=180^\circ-32^\circ-73^\circ=75^\circm∠ACB=180∘−32∘−73∘=75∘
The exterior measure at CCC is 180∘−75∘=105∘180^\circ-75^\circ=105^\circ180∘−75∘=105∘.
Triangle ABCABCABC has BCBCBC extended beyond CCC to DDD, with B,C,DB,C,DB,C,D in that order. If m∠ACD=121∘m\angle ACD=121^\circm∠ACD=121∘, m∠BAC=(3x+7)∘m\angle BAC=(3x+7)^\circm∠BAC=(3x+7)∘, and m∠ABC=(x+10)∘m\angle ABC=(x+10)^\circm∠ABC=(x+10)∘, what is m∠BACm\angle BACm∠BAC?
Correct answer: C
The remote angles add to 121∘121^\circ121∘.
(3x+7)+(x+10)=121(3x+7)+(x+10)=121(3x+7)+(x+10)=121
This gives 4x+17=1214x+17=1214x+17=121, then 4x=1044x=1044x=104 and x=26x=26x=26. Substitute into the requested measure: 3(26)+7=853(26)+7=853(26)+7=85, so m∠BAC=85∘m\angle BAC=85^\circm∠BAC=85∘. The other remote is 36∘36^\circ36∘, and the adjacent interior is 59∘59^\circ59∘.
A measurement record for a triangle gives an exterior angle of 87∘87^\circ87∘ and one of its remote interior angles as 92∘92^\circ92∘. Which conclusion follows?
The exterior angle theorem would give the other remote measure as
87∘−92∘=−5∘.87^\circ-92^\circ=-5^\circ.87∘−92∘=−5∘.
A triangle's interior measures are positive. Equivalently, the exterior angle inequality requires the exterior to exceed each remote angle, which this record violates.
At a triangle vertex, a non-reflex exterior angle has three times the measure of its adjacent interior angle. What is the exterior measure?
Let the adjacent interior measure be ccc, so the exterior measure is 3c3c3c. They form a linear pair.
c+3c=180∘c+3c=180^\circc+3c=180∘
Thus c=45∘c=45^\circc=45∘ and the exterior measure is 3(45∘)=135∘3(45^\circ)=135^\circ3(45∘)=135∘.
Triangle ABCABCABC has BCBCBC extended beyond CCC to DDD, with B,C,DB,C,DB,C,D in that order. The remote interior angles at AAA and BBB measure 54∘54^\circ54∘ and 62∘62^\circ62∘. Ray CECECE lies strictly inside ∠ACD\angle ACD∠ACD, and m∠ECD=39∘m\angle ECD=39^\circm∠ECD=39∘. What is m∠ACEm\angle ACEm∠ACE?
First the exterior angle theorem gives m∠ACD=54∘+62∘=116∘m\angle ACD=54^\circ+62^\circ=116^\circm∠ACD=54∘+62∘=116∘.
Because CECECE is inside the exterior angle, angle addition gives m∠ACE+m∠ECD=m∠ACDm\angle ACE+m\angle ECD=m\angle ACDm∠ACE+m∠ECD=m∠ACD.
m∠ACE=116∘−39∘=77∘m\angle ACE=116^\circ-39^\circ=77^\circm∠ACE=116∘−39∘=77∘
A path follows a triangle once, with all three non-reflex exterior turns to the right. Their degree measures are x+65x+65x+65, 2x+302x+302x+30, and 3x−53x-53x−5. What is xxx?
The consistently oriented exterior turns total 360∘360^\circ360∘.
(x+65)+(2x+30)+(3x−5)=360(x+65)+(2x+30)+(3x-5)=360(x+65)+(2x+30)+(3x−5)=360
This simplifies to 6x+90=3606x+90=3606x+90=360, so 6x=2706x=2706x=270 and x=45x=45x=45. The turns become 110∘110^\circ110∘, 120∘120^\circ120∘, and 130∘130^\circ130∘, all non-reflex. Their adjacent interiors are positive and total 180∘180^\circ180∘.
A path follows triangle ABCABCABC once, taking one non-reflex left exterior turn at each vertex. Its interior measures are a,b,ca,b,ca,b,c, and the corresponding exterior turn measures are eA,eB,eCe_A,e_B,e_CeA,eB,eC. Which step justifies a total exterior turn of 360∘360^\circ360∘?
Each exterior turn supplements the interior angle at its own vertex. Let TTT be the sum of the three exterior turn measures. Adding the three linear-pair equations gives
T=540∘−(a+b+c).T=540^\circ-(a+b+c).T=540∘−(a+b+c).
The triangle sum supplies a+b+c=180∘a+b+c=180^\circa+b+c=180∘. Substitution gives
T=540∘−180∘.T=540^\circ-180^\circ.T=540∘−180∘.
Therefore T=360∘T=360^\circT=360∘. The consistent left turns make these three exterior angles the turns of the stated traversal.
In triangle ABCABCABC, extend BCBCBC beyond CCC to DDD and extend ACACAC beyond CCC to EEE. Thus B,C,DB,C,DB,C,D and A,C,EA,C,EA,C,E are each in the stated order on distinct lines. If m∠ACB=72∘m\angle ACB=72^\circm∠ACB=72∘, what is m∠BCEm\angle BCEm∠BCE?
The exterior angle ∠BCE\angle BCE∠BCE shares side CBCBCB with the interior angle ∠BCA\angle BCA∠BCA, while CACACA and CECECE are opposite rays. They form a linear pair.
m∠BCE=180∘−72∘=108∘m\angle BCE=180^\circ-72^\circ=108^\circm∠BCE=180∘−72∘=108∘
It also equals the other exterior angle ∠ACD\angle ACD∠ACD, since those two are vertical angles formed by the distinct intersecting lines.
A triangle has an exterior angle of 104∘104^\circ104∘. One remote interior angle has measure aaa, and the other has measure 2a+8∘2a+8^\circ2a+8∘. What is the second remote measure?
The remote measures add to the exterior measure.
a+(2a+8∘)=104∘a+(2a+8^\circ)=104^\circa+(2a+8∘)=104∘
Thus 3a=96∘3a=96^\circ3a=96∘ and a=32∘a=32^\circa=32∘. The second remote measure is 2(32∘)+8∘=72∘2(32^\circ)+8^\circ=72^\circ2(32∘)+8∘=72∘. Their sum checks to 104∘104^\circ104∘.
A triangular path takes three non-reflex exterior turns, all to the left. Two measure 133∘133^\circ133∘ and 102∘102^\circ102∘. What is the interior angle at the third vertex?
The third exterior turn is the amount missing from the 360∘360^\circ360∘ total.
360∘−133∘−102∘=125∘360^\circ-133^\circ-102^\circ=125^\circ360∘−133∘−102∘=125∘
Its adjacent interior angle forms a linear pair, so its measure is 180∘−125∘=55∘180^\circ-125^\circ=55^\circ180∘−125∘=55∘.
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