Question

If m∠B = 111°, a = 12, and c = 6, what are the measures of the remaining side and angles? (it is a triangle) m∠A = 47.4°, m∠C = 21.6°, b = 15.2 m∠A = 47.4°, m∠C = 21.6°, b = 15.9 m∠A = 44.8°, m∠C = 24.2°, b = 15.9 m∠A = 44.8°, m∠C = 24.2°, b = 15.2

Answer 1

To answer, we will first need to use the Law of cosines. Assuming the angles ∠A, ∠B and ∠C are opposite to the sides a, b, and c, we can use the formula of the Law of Cosines to find the missing side:

`b^2=a^2+c^2-2ac\cos B^{}`

Thus, we have:

m∠B = 111°

a = 12

c = 6

`\begin{gathered} b^2=12^2+6^2-2\cdot12\cdot6\cdot\cos 111\degree \\ b^2=144+36-144\cdot(-0.35836\ldots) \\ b^2=231.604\ldots \\ b=15.2185\ldots\approx15.2 \end{gathered}`

Now that we know all sides and one of the angles, we can use the Law of Since:

`(\sin A)/(a)=(\sin B)/(b)=(\sin C)/(c)`

Thus:

`\begin{gathered} (\sin A)/(a)=(\sin B)/(b) \\ \sin A=a\cdot(\sin B)/(b)=12\cdot(\sin111\degree)/(15.2185\ldots)=(0.9335\ldots)/(15.2185\ldots)=0.7361\ldots \\ A=\arcsin 0.7361\ldots=47.4034\ldots\degree\approx47.4\degree \end{gathered}`

And:

`\begin{gathered} (\sin C)/(c)=(\sin B)/(b) \\ \sin C=c\cdot(\sin B)/(b)=6\cdot(\sin111\degree)/(15.2185\ldots)=(0.9335\ldots)/(15.2185\ldots)=0.3680\ldots \\ C=\arcsin 0.3680\ldots=21.5965\ldots\degree\approx21.6\degree \end{gathered}`

So, we have:

`\begin{gathered} m\angle A\approx47.4\degree \\ m\angle C\approx21.6\degree \\ b\approx15.2 \end{gathered}`

Which corresponds to the first alternative.