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

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`