Question

Solve the triangle.

A = 32°, a = 19, b = 14

Answer 1

first step is to draw a reasonable triangle

Answer 2

@$$
angle, side, side, use law of cosines

Answer 3

hmmm i would use the law of sines, because it is easier

Answer 4

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 5

notice that in the drawing, i made a 32 degree angle look somewhat like 32, maybe not exactly but somewhat
also it is important to note that \(a=19\) is longer than \(b=14\)

Answer 6

yeah sines dodges the ambiguous case

Answer 7

the via the law of sines you know
\[\frac{\sin(A)}{a}=\frac{\sin(B)}{b}\]so
\[\frac{\sin(32)}{19}=\frac{\sin(B)}{14}\] making
\[\sin(B)=\frac{14\sin(32)}{19}\]

Answer 8

you cannot use the law of cosines for this one
for the law of sines you need to know 3 out of the 4 numbers in \[\frac{\sin(A)}{a}=\frac{\sin(B)}{b}\] which you do

Answer 9

also it is the law of sines that may have an "ambiguous case" not the law of cosines
in any case once we have \[\sin(B)=\frac{14\sin(32)}{19}\] we get \(B\) via
\[B=\sin^{-1}\left(\frac{14\sin(32)}{19}\right)\]

Answer 10

i get \(B=22.98\)
http://www.wolframalpha.com/input/?i=arcsin%2814sin%2832%29%2F19%29

Answer 11

which i guess your answers are calling \(23\)

Answer 12

then you can easily find \(C\) since the angles have to add up to \(180\) and one is \(32\) the other \(23\)