Question

Solve the triangle.

A = 33°, a = 19, b = 14

Answer 1

I assume you've covered the "law of sines" already thus

Answer 2

i understand how to set everything up
19/sin33=14/sinB
19sinB=14sin33
sinB=14sin33/19
sinB=.4013
now what i dont understand is how to get it in degrees

Answer 3

\(\bf \cfrac{sin(A)}{a}=\cfrac{sin(B)}{b}\implies \cfrac{sin(33^o)}{19}=\cfrac{sin(B)}{14}
\\ \quad \\
\cfrac{sin(33^o)\cdot 14}{19}=sin(B)\implies sin^{-1}\left[\cfrac{sin(33^o)\cdot 14}{19}x\right]=sin^{-1}[sin(B)]
\\ \quad \\
sin^{-1}\left[\cfrac{sin(33^o)\cdot 14}{19}x\right]=\measuredangle B\)

Answer 4

hmmm I have an "x" hitchhiking there as a typo darn anyhow \(\bf \cfrac{sin(A)}{a}=\cfrac{sin(B)}{b}\implies \cfrac{sin(33^o)}{19}=\cfrac{sin(B)}{14}
\\ \quad \\
\cfrac{sin(33^o)\cdot 14}{19}=sin(B)\implies sin^{-1}\left[\cfrac{sin(33^o)\cdot 14}{19}\right]=sin^{-1}[sin(B)]
\\ \quad \\
sin^{-1}\left[\cfrac{sin(33^o)\cdot 14}{19}\right]=\measuredangle B\)

Answer 5

notice that your 33 are in degrees, thus make sure your calculator is in Degree mode

Answer 6

thank you now i understand how to get the value of angle B

Answer 7

yw

Answer 8

yes i have my mode set in degrees