Find Angle CAB ...?

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  • 5 months ago
    Favorite Answer

    apply law of sine to solve angle DCB

    ...BC............DB

    ---------- = ---------

    .sin(CDB)..sin(DCB)

    ..389............182

    ----------- = -----------

    .sin(82)......sin(DCB)

    389sin(DCB)  = 182sin(82)

    sin⁻¹(DCB) = 0.46

    DCB = 27.60°

    solving for angle DBC

    m∠DBC = 180 - 82 - 27.60 = 70.40°

    solving for Angle CAB

    ......CB.................AC

    --------------- = ---------------

    ..sin(CAB).........sin(CBA)

    ....389...................518

    --------------- = --------------

    ..sin(CAB)........sin(70.40)

    389sin(70.40) = 518sin(CAB)

    sin⁻¹(CAB) = 0.707

     CAB = 45.03° Answer//

  • 5 months ago

     In triangle BCD, BC = 389, BD = 182, and ∠ BDC = 82°.

     CD = 370.059  Angle ∠ B = 70.399° = 70°23'56″  Angle ∠ BCD = 27.601° = 27°36'4″  In triangle ACD, AC = 518, ∠ ADC = 98°  Angle ∠ A = 45.028° = 45°1'39″ 

  • david
    Lv 7
    5 months ago

    CB =  DB + CD  -  2CD*DBcosD  ...  CD = x

    389^2 = 182^2 + x^2 - 2*182*x*cos82

      CD = 370.059  ....  sorry I fell asleep and never came back .. 

  • Robert
    Lv 7
    5 months ago

    Step 1: By the law of sines,

    389 / sin 82 = 182 / sin ∠BCD;

    ∠BCD ≈ 27.6º

    So ∠CBD ≈ 70.4º

    Step 2: By the law of sines again,

    518 / sin 70.4 = 389 / sin ∠BAC;

    ∠BAC = 45º

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  • 5 months ago

    Applying the 'sine rule' on ∆CDB we have:

    389/sin82 = 182/sinC

    so, sinC = 182sin82/389

    Hence, C = 28°

    i.e. ∠ DCB = 28°

    Therefore, ∠ CBD = 70°

    Applying the 'sine rule' again to ∆CDB we have:

    CD/sin70 = 389/sin82

    so, CD = 389sin70/sin82 => 369 

    Applying the 'sine rule' to ∆ACD we have:

    518/sin98 = 369/sinA

    so, sinA = 369sin98/518

    Hence, ∠CAB = 45°

    This gives ∠ ACD = 37°

    i.e. ∠ ACB = 65°

    Checking for ∠ CAB + ∠ ABC + ∠ ACB gives: 

    45° + 70° + 65° = 180°

    :)>

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