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

1 Answer
Nov 3, 2016

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