Question

Find the solutions for a triangle with a = 16, c =12, and B = 63º.

a. A = 71.6°; C = 52.89; b =12.0

C. A = 68.6° C = 45.40; b=12.0

b. A = 71.6°; C = 45.40; b =15.0

d. A = 68.6°; C = 52.89; b=15.0

Answer 1

Answer:

b. A = 71.6°; C = 45.40°; b =15.0

Step-by-step explanation:

The missing values can be found with the help of the Law of Cosine and properties of triangles:

Side b (Law of Cosine)

$b = \sqrt{a^{2}+c^{2}-2\cdot a \cdot c \cdot \cos B}$

$b = \sqrt{16^{2}+12^{2}-2\cdot (16)\cdot (12) \cdot \cos 63^{\circ}}$

$b \approx 15.022$

Angle A (Law of Cosine)

$\cos A = -\frac{a^{2} - b^{2}-c^{2}}{2\cdot b \cdot c}$

$\cos A = - \frac{16^{2}-15.022^{2}-12^{2}}{2\cdot (15.022)\cdot (12)}$

$\cos A = 0.315$

$A= \cos^{-1} 0.315$

$A \approx 71.639^{\circ}$

Angle C (Sum of internal angles in triangles)

$C = 180^{\circ} - 63^{\circ} - 71.639^{\circ}$

$C = 45.361^{\circ}$

Hence, the right answer is B.