in triangle abc |BC|=4.12cm ,|AC|=6.82cm and angle BCA is 42.5 . Calculate|AB| . and angle bac and angle abc​

i. |AB| = 4.70 cm

ii. BAC =

iii. ABC =

Explanation:

i Let |AB| be represented by c, |AC| by b and |BC| by a. Since the question would give an included angle triangle, we first apply cosine rule.

= + - 2ab Cos C

   = + - 2(4.12*6.82) Cos

   = 16.9744 + 46.5124 - 56.1968*0.7373

   = 63.4868 - 41.4340

= 22.0528

c =

  = 4.6960

c = 4.70 cm

Thus |AB| = 4.70 cm

ii. Applying the sine rule;

= =

=

sin A =

         = 0.5922

A = 0.5922

   = 36.3

BAC =

But,

iii. BCA + BAC + ABC =

+ + ABC =

ABC = - 78.8

       = 101.2

ABC =

ok so u have2 basically

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