How do you solve the triangle given A=30, a=14, b=28?

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Answer 1 · 2018-04-25T14:55:28

It's yet another 30,60,90 triangle so #B=90^circ#, #C=60^circ# and #c=14 sqrt{3}#.

Oh my gosh, another trig problem with #30^circ#. Why have we created an entire field that handles only two triangles?

#a/sinA = b /sin B = c / sin C#

# sin B = b/a sin A = 28/14 sin 30^circ = 2 (1/2) = 1 #

Interestingly, #1# is the only sine that uniquely determines a triangle angle because it's its own supplement.

# B = 90^ circ #

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^circ#

# c = a \ sin C/sin A = {(14) (sqrt{3}/2)} / (1/2) = 14 sqrt{3} #