Question

Solve the triangle. Round answers to two decimal places. Enter the degree symbol by

typing deg.

Answer 1

hmmm let's find hmmm say angle who knows, ∡C, using the law of cosines.

`\textit{Law of Cosines}\\\\ \cfrac{a^2+b^2-c^2}{2ab}=\cos(C)\implies \cos^(-1)\left(\cfrac{a^2+b^2-c^2}{2ab}\right)=\measuredangle C \\\\[-0.35em] ~\dotfill\\\\ \cos^(-1)\left(\cfrac{18^2+14^2-12^2}{2(18)(14)}\right)=\measuredangle C \implies \cos^(-1)\left(\cfrac{ 376 }{ 504 }\right)=\measuredangle C \\\\\\ \cos^(-1)\left(\cfrac{ 47 }{ 63 }\right)=\measuredangle C\implies \cos^(-1)(0.7460317) \approx \measuredangle C \implies 41.75^o \approx \measuredangle C`

well then, let's now us the law of sines to get hmm say ∡A

`\textit{Law of Sines} \\\\ \cfrac{\sin(\measuredangle A)}{a}=\cfrac{\sin(\measuredangle B)}{b}=\cfrac{\sin(\measuredangle C)}{c} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{\sin( A )}{18}\approx\cfrac{\sin( 41.75^o )}{12}\implies 12\sin(A)\approx18\sin(41.75^o) \implies \sin(A)\approx\cfrac{18\sin(41.75^o)}{12} \\\\\\ A\approx\sin^(-1)\left( ~~ \cfrac{18\sin( 41.75^o)}{12} ~~\right)\implies A\approx 87.22^o`

well, if we subtract ∡C and ∡A from the triangle's interior angle sum, we'll get that ∡B ≈ 51.03°.