Question

Solve the triangle. A = 32°, a = 19, b = 14 (1 point) B = 23°, C = 125°, c ≈17.6 Cannot be solved B = 23°, C = 125°, c ≈29.4 B = 23°, C = 145°, c ≈23.5

Answer 1

Answer:

The answer is below

Step-by-step explanation:

Given the triangle with: A = 32°, a = 19, b = 14

The sine rule states that for a triangle with lengths of a, b and c and the corresponding angles which are opposite the sides as A, B and C, then the following rule holds:

$\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$

Given, that for triangle ABC; A = 32°, a = 19, b = 14. therefore:

$\frac{a}{sinA}=\frac{b}{sinB}\\\\\frac{19}{sin(32)}=\frac{14}{sin(B)}\\\\sin(B)=\frac{14*sin(32)}{19} \\\\sin(B)=0.39\\\\B=sin^{-1}(0.39)\\\\B=23^o$

A + B + C = 180° (sum of angles in a triangle)

32 + 23 + C = 180

55 + C = 180

C = 180 - 55

C = 125°

$\frac{a}{sin(A)}=\frac{c}{sin(C)}\\\\\frac{19}{sin(32)}=\frac{c}{sin(125)} \\\\c=\frac{19*sin(125)}{sin(32)} \\\\c=29.4$

Answer 2

Answer:

B = 23°, C = 125°, c ≈29.4

Step-by-step explanation: