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:

$a:sinalpha=c:singamma$

$10:sin30=c:sin105$

$c=10 sin 105/sin30$

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

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