Question

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

Answer 1

Answer:

i. |AB| = 4.70 cm

ii. BAC = $36.3^{o}$

iii. ABC = $101.2^{o}$

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.

$c^{2}$ = $a^{2}$ + $b^{2}$ - 2ab Cos C

   = $(4.12)^{2}$ + $(6.82)^{2}$ - 2(4.12*6.82) Cos $42.5^{o}$

   = 16.9744 + 46.5124 - 56.1968*0.7373

   = 63.4868 - 41.4340

$c^{2}$ = 22.0528

c = $\sqrt{22.0528}$

  = 4.6960

c = 4.70 cm

Thus |AB| = 4.70 cm

ii. Applying the sine rule;

$\frac{a}{sin A}$ = $\frac{b}{sin B}$ = $\frac{c}{sin C}$

$\frac{4.12}{Sin A}$ = $\frac{4.7}{sin 42.5}$

sin A = $\frac{4.12*sin 42.5}{4.7}$

         = 0.5922

A = $sin^{-1}$ 0.5922

   = 36.3

BAC = $36.3^{o}$

But,

iii. BCA + BAC + ABC = $180^{o}$

$42.5^{o}$ + $36.3^{o}$ + ABC = $180^{o}$

ABC = $180^{o}$ - 78.8

       = 101.2

ABC = $101.2^{o}$