12 multiple-choice questions, progressively harder.
In triangle ABCABCABC, point DDD is between AAA and BBB. Given m∠CAB=33∘m\angle CAB=33^\circm∠CAB=33∘, m∠ABC=59∘m\angle ABC=59^\circm∠ABC=59∘, and m∠ACD=41∘m\angle ACD=41^\circm∠ACD=41∘, find m∠DCBm\angle DCBm∠DCB.
Solution
Correct answer: D
In triangle ACDACDACD, rays ADADAD and ABABAB agree, so its angle at AAA is 33∘33^\circ33∘.
m∠ADC=180∘−33∘−41∘=106∘.m\angle ADC=180^\circ-33^\circ-41^\circ=106^\circ.m∠ADC=180∘−33∘−41∘=106∘.
At DDD, rays DADADA and DBDBDB are opposite, giving a linear pair.
m∠CDB=180∘−106∘=74∘.m\angle CDB=180^\circ-106^\circ=74^\circ.m∠CDB=180∘−106∘=74∘.
Triangle CDBCDBCDB has 59∘59^\circ59∘ at BBB, so m∠DCB=180∘−74∘−59∘=47∘.m\angle DCB=180^\circ-74^\circ-59^\circ=47^\circ.m∠DCB=180∘−74∘−59∘=47∘.
In triangle ABCABCABC, DDD lies inside side ABABAB, EEE lies inside side ACACAC, and DE∥BCDE\parallel BCDE∥BC. The angles at A,BA,BA,B measure 57∘,48∘57^\circ,48^\circ57∘,48∘. Find m∠AEDm\angle AEDm∠AED.
Correct answer: B
Because DE∥BCDE\parallel BCDE∥BC, transversal ABABAB gives corresponding angles ∠ADE≅∠ABC\angle ADE\cong\angle ABC∠ADE≅∠ABC. Thus m∠ADE=48∘m\angle ADE=48^\circm∠ADE=48∘.
Rays AD,ABAD,ABAD,AB agree and rays AE,ACAE,ACAE,AC agree, so the angle at AAA in triangle ADEADEADE is 57∘57^\circ57∘. Its angle sum gives
m∠AED=180∘−57∘−48∘=75∘.m\angle AED=180^\circ-57^\circ-48^\circ=75^\circ.m∠AED=180∘−57∘−48∘=75∘.
A triangle has angle measures (2x+1)∘(2x+1)^\circ(2x+1)∘, (4x−3)∘(4x-3)^\circ(4x−3)∘, and (5x+6)∘(5x+6)^\circ(5x+6)∘. Which describes the triangle after solving for xxx?
Correct answer: A
The three variable terms total 11x11x11x, and the constant terms total 444. The triangle sum therefore gives 11x+4=180.11x+4=180.11x+4=180.
Subtract 444 and divide by 111111 to obtain x=16x=16x=16. Substitution gives angle measures 33∘,61∘,86∘33^\circ,61^\circ,86^\circ33∘,61∘,86∘.
Their sum is 180∘180^\circ180∘ and each is less than 90∘90^\circ90∘, so the triangle is acute with largest angle 86∘86^\circ86∘.
The acute angles of a right triangle differ by 26∘26^\circ26∘. Find the larger one.
Correct answer: C
Let the smaller measure be vvv degrees. The larger is v+26v+26v+26 degrees, and the two acute measures total 90∘90^\circ90∘.
v+(v+26)=90.v+(v+26)=90.v+(v+26)=90.
Then 2v=642v=642v=64 and v=32v=32v=32. The larger measure is 32∘+26∘=58∘32^\circ+26^\circ=58^\circ32∘+26∘=58∘.
Two angles of a nondegenerate triangle are complementary. Which argument establishes its angle classification?
Complementary means the two measures total 90∘90^\circ90∘. The triangle sum leaves 180∘−90∘=90∘.180^\circ-90^\circ=90^\circ.180∘−90∘=90∘.
So the remaining angle is right, and the triangle is right. The two complementary angles need not be congruent.
In the triangle angle-sum construction, D,C,ED,C,ED,C,E are collinear in that order, DE∥ABDE\parallel ABDE∥AB, and the triangle is below DEDEDE as shown. Which two angle congruences follow by alternate interior angles?
Transversal ACACAC connects ∠DCA\angle DCA∠DCA with ∠CAB\angle CAB∠CAB: the angles lie inside the parallels on opposite sides of that transversal.
Transversal BCBCBC similarly connects ∠BCE\angle BCE∠BCE with ∠ABC\angle ABC∠ABC. Equal base angles or equal top pieces are not part of the givens.
Triangle ABCABCABC is right at BBB. Point DDD lies inside side ACACAC, and BD⊥ACBD\perp ACBD⊥AC. If m∠BAC=28∘m\angle BAC=28^\circm∠BAC=28∘, find m∠DBCm\angle DBCm∠DBC.
In the whole right triangle, m∠ACB=90∘−28∘=62∘.m\angle ACB=90^\circ-28^\circ=62^\circ.m∠ACB=90∘−28∘=62∘.
In triangle BDCBDCBDC, the angle at DDD is 90∘90^\circ90∘ because BD⊥ACBD\perp ACBD⊥AC. Its angle at CCC is the same 62∘62^\circ62∘ angle, since rays CD,CACD,CACD,CA agree.
m∠DBC=180∘−90∘−62∘=28∘.m\angle DBC=180^\circ-90^\circ-62^\circ=28^\circ.m∠DBC=180∘−90∘−62∘=28∘.
In triangle ABCABCABC, DDD lies between BBB and CCC, and ADADAD bisects ∠BAC\angle BAC∠BAC. The angles at B,CB,CB,C measure 58∘,66∘58^\circ,66^\circ58∘,66∘. Find m∠ADCm\angle ADCm∠ADC.
The whole angle at AAA is 180∘−58∘−66∘=56∘.180^\circ-58^\circ-66^\circ=56^\circ.180∘−58∘−66∘=56∘.
The bisector divides it into two equal 28∘28^\circ28∘ angles. In triangle ADCADCADC, the angle at CCC is 66∘66^\circ66∘ because DDD lies on BCBCBC.
m∠ADC=180∘−28∘−66∘=86∘.m\angle ADC=180^\circ-28^\circ-66^\circ=86^\circ.m∠ADC=180∘−28∘−66∘=86∘.
Two distinct lines contain A,O,CA,O,CA,O,C in that order and B,O,DB,O,DB,O,D in that order. In triangle AOBAOBAOB, the angles at A,BA,BA,B measure (2x+3)∘,(3x+7)∘(2x+3)^\circ,(3x+7)^\circ(2x+3)∘,(3x+7)∘. In triangle CODCODCOD, the angles at C,DC,DC,D measure 44∘,56∘44^\circ,56^\circ44∘,56∘. Find m∠OABm\angle OABm∠OAB.
Triangle CODCODCOD has m∠COD=180∘−44∘−56∘=80∘.m\angle COD=180^\circ-44^\circ-56^\circ=80^\circ.m∠COD=180∘−44∘−56∘=80∘.
Vertical angles give m∠AOB=80∘m\angle AOB=80^\circm∠AOB=80∘. Apply the triangle sum to AOBAOBAOB.
(2x+3)+(3x+7)+80=180.(2x+3)+(3x+7)+80=180.(2x+3)+(3x+7)+80=180.
This gives 5x=905x=905x=90, so x=18x=18x=18. The requested angle is (2(18)+3)∘=39∘(2(18)+3)^\circ=39^\circ(2(18)+3)∘=39∘.
Two interior angles of a triangle are proposed as (2u+4)∘(2u+4)^\circ(2u+4)∘ and (u+8)∘(u+8)^\circ(u+8)∘. Which value of uuu gives two positive measures whose sum leaves a positive third angle?
At u=55u=55u=55, the proposed measures are 114∘114^\circ114∘ and 63∘63^\circ63∘. They leave 180∘−114∘−63∘=3∘,180^\circ-114^\circ-63^\circ=3^\circ,180∘−114∘−63∘=3∘, which is positive.
At u=−5u=-5u=−5 or u=−3u=-3u=−3, the first expression is negative. At u=57u=57u=57, the two measures total 183∘183^\circ183∘, leaving a negative third measure. Only 555555 passes all the stated checks.
Triangles ABCABCABC and PQRPQRPQR satisfy ∠A≅∠Q\angle A\cong\angle Q∠A≅∠Q and ∠B≅∠R\angle B\cong\angle R∠B≅∠R. Which remaining correspondence follows?
The two given pairs account for angles A,BA,BA,B in the first triangle and Q,RQ,RQ,R in the second.
Subtracting those matching measures from 180∘180^\circ180∘ leaves equal measures at the unused vertices C,PC,PC,P. The third-angle theorem gives ∠C≅∠P\angle C\cong\angle P∠C≅∠P.
In triangle KLMKLMKLM, NNN lies between KKK and LLL. Given m∠LKM=36∘m\angle LKM=36^\circm∠LKM=36∘, m∠KLM=72∘m\angle KLM=72^\circm∠KLM=72∘, and m∠KMN=24∘m\angle KMN=24^\circm∠KMN=24∘, which sequence correctly finds m∠NMLm\angle NMLm∠NML through the smaller triangles?
Triangle KMNKMNKMN has 36∘36^\circ36∘ at KKK and 24∘24^\circ24∘ at MMM, so m∠KNM=180∘−36∘−24∘=120∘.m\angle KNM=180^\circ-36^\circ-24^\circ=120^\circ.m∠KNM=180∘−36∘−24∘=120∘.
The angles at NNN form a linear pair, giving m∠MNL=60∘m\angle MNL=60^\circm∠MNL=60∘. In triangle MNLMNLMNL, the angle at LLL is 72∘72^\circ72∘, so m∠NML=180∘−60∘−72∘=48∘.m\angle NML=180^\circ-60^\circ-72^\circ=48^\circ.m∠NML=180∘−60∘−72∘=48∘.
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