Question

PLEASE HELP.find the solutions for a triangle with a =16 c =12 and b=63 degrees

Answer 1

A.16
b.12
c.63
try to have the triangle and draw a little drawling it will help
and start from there

Answer 2

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,

$b^2=a^2+c^2-2ab cos B$

By substituting the values,

$b^2 = 16^2+12^2-2\times 16\times 12 \times cos 63^{\circ}=256 + 144 - 384 cos 63^{\circ}=225.67$

$\implies b\approx 15.02$

By the law of sine,

$\frac{sin B}{AC}=\frac{sin C}{AB}$

$\frac{sin 63^{\circ}}{15.02}=\frac{sin C}{12}$

$\implies sin C= \frac{12sin 63^{\circ}}{15.02}= 0.7119$

$\implies \angle C\approx 45.39^{\circ}$

∵ 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°