How do you solve the triangle when angle W = 58˚, w = 12 cm, and h = 14 cm?

1 Answer
Mar 10, 2018

I'm going to make a table of all our information:

#color(white)(.) Len g t h color(white)(.) | color(white)(.) Ang l e#

#w = 12 color(white)(....) | color(white)(.) W = 58#
#h = 14 color(white)(. ...) | color(white)(.) H = color(green)(?)#
#x = color(green)(?)color(white)(4) color(white)(....) | X = color(green)(?)#

We'll use #SinA/a = SinB/b#

#Sin (58)/12 = Sin H/14#

#0.071 = SinH/14#

#0.989 = SinH#

#H = Sin^-1(0.989)#

#H = 81.65#

Now our table is

#w = 12 color(white)(....) | color(white)(.) W = 58#
#h = 14 color(white)(. ...) | color(white)(.) H = 81.65#
#x = color(green)(?)color(white)(4) color(white)(....) | X = color(green)(?)#

Since we know a triangle has #180# total degree, we can solve for the last angle, #X#

#180 - 81.65 - 58 = 40.35#

#w = 12 color(white)(....) | color(white)(.) W = 58#
#h = 14 color(white)(. ...) | color(white)(.) H = 81.65#
#x = color(green)(?)color(white)(4) color(white)(....) | X = 40.35#

Now we can use #SinA/a = SinB/b# again

#Sin58/12 = Sin40.35/x#

#0.071 = Sin40.35/x#

#x = Sin(40.35)/0.071#

#x = 9.16#

Now our table is full

#w = 12 color(white)(....) | color(white)(.) W = 58#
#h = 14 color(white)(. ...) | color(white)(.) H = 81.65#
#x = 9.16 color(white)(.) | X = 40.35#