Oh my gosh, another trig problem with 30^circ30∘. Why have we created an entire field that handles only two triangles?
The Law of Sines says
a/sinA = b /sin B = c / sin CasinA=bsinB=csinC
sin B = b/a sin A = 28/14 sin 30^circ = 2 (1/2) = 1 sinB=basinA=2814sin30∘=2(12)=1
Interestingly, 11 is the only sine that uniquely determines a triangle angle because it's its own supplement.
B = 90^ circ B=90∘
It's another 30,60,90 triangle and I just want to crawl in a hole. Get a new triangle, question writers!
C=180^circ -A - B = 180^circ - 30^circ - 90^circ = 60^circC=180∘−A−B=180∘−30∘−90∘=60∘
c = a \ sin C/sin A = {(14) (sqrt{3}/2)} / (1/2) = 14 sqrt{3} c=a sinCsinA=(14)(√32)12=14√3
