Sunday, January 18, 2015

Geometry Problem 1074: Quadrilateral, Right Triangle, Angles, 15, 22,5, 30, 90 Degrees

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below to enlarge it.

Online Math: Geometry Problem 1074: Quadrilateral, Right Triangle, Angles, 15, 22,5, 30, 90 Degrees.

6 comments:

  1. Let O is a circle through A, B and D where cuts AC and DC at E and F respectively.
    Let AD and BC intersect at G
    Note that A, B, E, F and D concyclic
    Because ∠GBA an exterior angle of triangle ABC then m ∠GBA = 52.5 degree
    so that m∠DBC = 105 degree. This implies ∠DBC = 30 degree
    Because BADF cyclic then m∠BAF = 30 degree so that m∠EAF = 7.5 degree = m∠EDF (A, B, E, F and D concyclic)
    Because ∠AED an exterior angle of triangle ECD then m ∠AED = (7.5 + 15) degree = 22.5 degree = x
    (From wisna email: putu_wisna_ariawan@yahoo.com)

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  2. To Anonymous
    Refer to line 5 "so that m∠DBC = 105 degree"
    Please provide more explanation how do you get 105 degree.
    http://s17.postimg.org/mmbwojrcv/pro_1074.png

    Peter Tran

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  3. To Peter Tran
    I am sory because there was an error in my solution in line 5.
    Thanks for your corection. But I am sure that x = 22.5 degree.
    I still working to find that solution.

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  4. With out loss of generality assume AB = 1.
    (all angles are measured in degrees)
    First note CAD = 67.5,
    ADC = (180 - 82.5),
    ABC = (180 - 57.5)
    By Sine Rule applied to Δ ABC:
    AC / sin 57.5 = AB / sin 30 = 2.
    So AC = 2 sin 57.5 = 2 cos 37.5 ..... (i).
    By Sine Rule applied to ΔADC:
    AD / sin 15 = AC / sin 82.5 = AC / cos 7.5
    So AD = AC. [sin 15 / cos 7.5]
    = 2 [cos 37.5].[2 sin 7.5]
    = 2[2 cos 37.5sin 7.5]
    = 2[sin 45 – sin 30]
    = 2[1/(sqrt 2) - 1/2]
    = (sqrt 2) - 1 = tan 22.5
    So tan x = AD / AB = (tan 22.5) / 1 = tan 22.5
    Hence x = 22.5 degrees

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  5. Let O be the center of a circle through B, D and C where cuts AC at P so we have triangle BPO equilateral (I) ( Because m∠BOP = 2m∠BCP=30º.Now have m∠BOD = 2.m∠BCD = 90º so #BODA is cyclic because m∠BAD + m∠BOD = 90º+90º= 180º then m∠BAO = m∠BDO = 45º but m∠BAO = m∠BAC+m∠CAO then m∠CAO = 22.5º (II). Circunference ABOD cuts AC at Q but AC bissects ∠BAO then we have BQ = QO then using (I), triangles BPQ and QPO are congruents by equal sides then m∠BPQ = 30º = m∠BAP + m∠ABP ( ∠BPQ external angle) .: m∠ABP = 30º - 22.5º = 7.5º (III) but m∠PBD = m∠PCD = 15º (inscritive angles), finaly using (III) we have m∠ABD = m∠ABP + m∠PBD = 7.5º + 15º .: m∠ABD = 22.5 º

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  6. Let O be the circumcentre of ABC and let CO and AD meet at E.

    < AOB = 60 so Tr. OAB is equilateral. Also < BOC = 45 so < AOE = 75. Hence <OAC = OCA = 37.5. Hence < DCE = 22.5 and < OAE = 30 so it follows that AO = AE

    Therefore Tr. BAE is right isoceles and so < AEB = 45 = < DCB implying that BCED is cyclic.

    So < DBE = < DCE = 22.5 and so x = 22.5

    Sumith Peiris
    Moratuwa
    Sri Lanka

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