# How do you solve the triangle given A= 30 degrees, b= 8.9 mi c= 6.0 mi?

May 2, 2018

Here's the situation

Let's put this in a table:

color(white)(..)A n g l e color(white)(. . . . ?) | color(white)(..) L e n g t h
color(white)(..) A = 30^o color(white)(..?) | color(white)(..) a =?
color(white)(..) B = ? color(white)(30^o)color(white)(..) | color(white)(..) b =8.9
color(white)(..) C = ? color(white)(30^o)color(white)(..) | color(white)(..) c =6.0

Law of Cosines

${a}^{2} = {b}^{2} + {c}^{2} - 2 \times b \times c \times \cos \left(A\right)$

$a = \sqrt{{b}^{2} + {c}^{2} - 2 \times b \times c \times \cos \left(A\right)}$

$a = \sqrt{{8.9}^{2} + {6}^{2} - 2 \times 8.9 \times 6 \times \cos \left(30\right)}$

$a = \sqrt{79.21 + 36 - 92.5}$

$a = 4.77$

color(white)(..)A n g l e color(white)(. . . . ?) | color(white)(..) L e n g t h
color(white)(..) A = 30^o color(white)(..?) | color(white)(..) a = 4.77
color(white)(..) B = ? color(white)(30^o)color(white)(..) | color(white)(..) b =8.9
color(white)(..) C = ? color(white)(30^o)color(white)(..) | color(white)(..) c =6.0

Law of Sine

$\sin \frac{A}{a} = \sin \frac{C}{c}$

${\sin}^{- 1} \left(c \times \sin \frac{A}{a}\right) = C$

${\sin}^{- 1} \left(6.0 \times \sin \frac{30}{4.77}\right) = C = {39}^{o}$

color(white)(..)A n g l e color(white)(. . . . ?) | color(white)(..) L e n g t h
color(white)(..) A = 30^o color(white)(..?) | color(white)(..) a = 4.77
color(white)(..) B = ? color(white)(30^o)color(white)(..) | color(white)(..) b =8.9
color(white)(..) C = 39^o color(white)(..?) | color(white)(..) c =6.0

Since the angle of a triangle must add up to ${180}^{o}$, we already know the angle of $B$:

$180 - 30 - 39 = 111$

color(white)(..)A n g l e color(white)(. . . . ?1) | color(white)(..) L e n g t h
color(white)(..) A = 30^o color(white)(..?1) | color(white)(..) a = 4.77
color(white)(..) B = 111^o color(white)(. .?) | color(white)(..) b =8.9
color(white)(..) C = 39^o color(white)(. .? 1) | color(white)(..) c =6.0