# Find Angle CAB ...?

Relevance
• 5 months ago

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

• 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

• 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°

:)>