How do you solve a triangle ABC given A= 30 degrees, B= 45 degrees, a= 10?

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Answer 1 · 2016-06-19T08:28:35

#C=105 degrees#; #b=10sqrt(2)#; #c=5(sqrt(2)+sqrt(6))#

First calculate the third angle by subracting:

#C=180-(30+45)=105#

Then, you know that in any triangle is true that (Euler theorem)

#a:sinalpha=b:sinbeta=c:singamma#

so use:

#a:sinalpha=b:sinbeta#

#10:sin30=b:sin45#

#b=10sin45/sin30#

#b=10(sqrt(2)/2)/(1/2)#

#b=10sqrt(2)#

Again use:

#a:sinalpha=c:singamma#

#10:sin30=c:sin105#

#c=10 sin 105/sin30#

#c=10((sqrt(2)+sqrt(6))/4)/(1/2)#

#c=5(sqrt(2)+sqrt(6))#