Answer:
Side b is 15.02,
Angle C is 45.39° and angle A is 71.61°
Step-by-step explanation:
Let in triangle ABC,
BC = a = 16 unit,
AB = c = 12 unit,
∠B = 63°,
By the law of cosine,
[tex]b^2=a^2+c^2-2ab cos B[/tex]
By substituting the values,
[tex]b^2 = 16^2+12^2-2\times 16\times 12 \times cos 63^{\circ}=256 + 144 - 384 cos 63^{\circ}=225.67[/tex]
[tex]\implies b\approx 15.02[/tex]
By the law of sine,
[tex]\frac{sin B}{AC}=\frac{sin C}{AB}[/tex]
[tex]\frac{sin 63^{\circ}}{15.02}=\frac{sin C}{12}[/tex]
[tex]\implies sin C= \frac{12sin 63^{\circ}}{15.02}= 0.7119[/tex]
[tex]\implies \angle C\approx 45.39^{\circ}[/tex]
∵ The sum of all interior angles of a triangle is supplementary,
⇒ ∠A + ∠B + ∠C = 180°
⇒ ∠A + 63° + 45.39° = 180°
⇒ ∠ A + 108.39° = 180°
⇒ ∠A = 71.61°
