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xz_007 [3.2K]
4 years ago
11

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

Mathematics
2 answers:
givi [52]4 years ago
7 0
A.16
b.12
c.63
try to have the triangle and draw a little drawling it will help
and start from there
nalin [4]4 years ago
5 0

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°

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