Question

How do you determine whether `triangle ABC` has no, one, or two solutions given `A=95^circ, a=19, b=12`?

Answer 1

one soltion only

`B=39.0^0, C=46.0^0, c=13.7`

Explanation:

The triangle for this problem is:

NTS

The Sine rule

`a/(sinA)=b/(sinB)=c/(sinC)`

may be used ot find either a side OR an angle if one side and an opposite angle are know.

In this case we need to find an angle first

`19/(sin95)=12/(sinB)=c/(sinC)`

we have to find `sinB` first

`19/(sin95)=12/(sinB)`

`=>sin95/19=sinB/12`

ie

`sinB=(12xxsin95)/19`

`sinB=0.6291755988`

`=>B=38.98932587^0`

`C=180-(95+38.98932587)`

`C=46.01067413^0`

to find the missing side

`c/sinC=19/sin95`

`c=(19xxsinC)/sin95`

#c=13.72213169

so the solution is , all given to 1dp

`B=39.0^0, C=46.0^0, c=13.7`

with the sine rule there is always the possibility of a second set of solutions for the triangle--the AMBIGUOUS case.

to check we take the first angle found

`B=39.9`

subtract from `180^0, because sin(180-x)=sinx`

`B'=180-39.9=140.1`

add this to A

`=95+140.1>180, :.`no second solution