Question

How do you solve the triangle ABC, given b=40, B=45, c=15?

Answer 1

`angleA=120^circ`
`angleC~~15^circ`
`a~~49`

Explanation:

This is an SSA triangle so you can use the Law of Sine to solve this
Law of Sine = `sinA/a=sinB/b=sinC/c`
in this case we use `sinB/c=sinC/c`
Given: `b=40, B=45^circ, c=15`

first let's find `angleC`
`sin45^circ/40=sinC/15`

`15sin45^circ=40sinC`

`angleC=sin^-1((15sin45^circ)/40)~~ 15^circ`
`angleC~~15^circ`

now we have `angleB` and `angleC` so we can subtract both from
`180^circ` to find the third angle
`angleA=180^circ-(15^circ+45^circ)=120^circ`
`angleA=120^circ`

to find `a` we use Law of Sine again
`sin45^circ/40=sin120^circ/a`
`asin45^circ=40sin120^circ`
`a=(40sin120^circ)/(sin45^circ)~~49`
`a~~49`