Question

How do you solve the triangle given `trianglePQR, p=12, mangleQ=80, mangleR=30`?

Answer 1

`Pcolor(white)(.)70^o color(white)(.............)pcolor(white)(.)12`
`Qcolor(white)(.)80^o color(white)(.............)qcolor(white)(.)12.576`
`Rcolor(white)(.)30^o color(white)(.............)rcolor(white)(.)6.385`

Explanation:

This si a table of what we know:

I like to put all the information in one place so I can keep track of everything
`Pcolor(white)(.) ?color(white)(.............)pcolor(white)(.)12`
`Qcolor(white)(.)80 color(white)(.............)qcolor(white)(.)?`
`Rcolor(white)(.)30 color(white)(.............)rcolor(white)(.)?`

Capital letters are angles, lowercase are lengths

We know that all angles in a triangle must add to `180^o`. So, if we have `80^o` and `30^o` (`110^o`), then the last remaining angle must be `70^o` `(180-110=70)`.

`Pcolor(white)(.)70^o color(white)(.............)pcolor(white)(.)12`
`Qcolor(white)(.)80^o color(white)(.............)qcolor(white)(.)?`
`Rcolor(white)(.)30^o color(white)(.............)rcolor(white)(.)?`

Now we know one angle-length pair (Q), so we can find the remaining lengths.

`color(white)(0)`
Solving for `color(red)(q)`
`(sin(70))/12=(sin(80))/q`

`sin(80)*12/(sin(70))=q`

`q~~12.576`

`color(white)(0)`

Solving for `color(red)(r)`

`(sin(70))/12=(sin(30))/r`

`sin(30)*12/(sin(70))=r`

`r~~6.385`

Now we have everything:

`Pcolor(white)(.)70 color(white)(.............)pcolor(white)(.)12`
`Qcolor(white)(.)80 color(white)(.............)qcolor(white)(.)12.576`
`Rcolor(white)(.)30 color(white)(.............)rcolor(white)(.)6.385`