Can someone help me solve for this angle?

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Answer 1 · 2018-03-01T02:43:29

$78.8^circ$

From the cosine law for triangles, we have

$a^2 = b^2+c^2-2bc cos/_A$

so that

$a^2 = 30^2+8^2-2times 30 times 8 cos 86^\circ ~~ 930.5 implies a ~~ 30.5$

Again

$b^2=a^2+c^2-2ac cos/_B implies cos/_B = {a^2+c^2-b^2}/{2ac}$

so that

$cos/_B = {30.5^2+8^2-30^2}/{2times 30.5 times 8} ~~ 0.194 implies$
$/_B = cos^-1 0.194 = 78.8^\circ$