Answer:
i. |AB| = 4.70 cm
ii. BAC = [tex]36.3^{o}[/tex]
iii. ABC = [tex]101.2^{o}[/tex]
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.
[tex]c^{2}[/tex] = [tex]a^{2}[/tex] + [tex]b^{2}[/tex] - 2ab Cos C
= [tex](4.12)^{2}[/tex] + [tex](6.82)^{2}[/tex] - 2(4.12*6.82) Cos [tex]42.5^{o}[/tex]
= 16.9744 + 46.5124 - 56.1968*0.7373
= 63.4868 - 41.4340
[tex]c^{2}[/tex] = 22.0528
c = [tex]\sqrt{22.0528}[/tex]
= 4.6960
c = 4.70 cm
Thus |AB| = 4.70 cm
ii. Applying the sine rule;
[tex]\frac{a}{sin A}[/tex] = [tex]\frac{b}{sin B}[/tex] = [tex]\frac{c}{sin C}[/tex]
[tex]\frac{4.12}{Sin A}[/tex] = [tex]\frac{4.7}{sin 42.5}[/tex]
sin A = [tex]\frac{4.12*sin 42.5}{4.7}[/tex]
= 0.5922
A = [tex]sin^{-1}[/tex] 0.5922
= 36.3
BAC = [tex]36.3^{o}[/tex]
But,
iii. BCA + BAC + ABC = [tex]180^{o}[/tex]
[tex]42.5^{o}[/tex] + [tex]36.3^{o}[/tex] + ABC = [tex]180^{o}[/tex]
ABC = [tex]180^{o}[/tex] - 78.8
= 101.2
ABC = [tex]101.2^{o}[/tex]
