Question

current attempt in progress
solve the triangle.
no triangle with the given measurements.

Explanation:

Step1: Apply the Law of Cosines

The Law of Cosines states that \(c^{2}=a^{2}+b^{2}-2ab\cos C\). But first, we can check the sum of angles. The sum of angles in a triangle is \(180^{\circ}\). Let's assume the side - angle relations.
Another approach is using the Law of Sines \(\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\). Let's assume the given side \(c = 5.0\) m, side \(4.6\) m and angle \(70^{\circ}\).
If we use the Law of Sines: \(\frac{\sin70^{\circ}}{5}=\frac{\sin\beta}{4.6}\)

$$ \sin\beta=\frac{4.6\sin70^{\circ}}{5} $$

$$ \sin\beta=\frac{4.6\times0.9397}{5}=\frac{4.3226}{5}=0.8645 $$

\(\beta=\sin^{- 1}(0.8645)\approx60^{\circ}\)
Then \(\alpha=180^{\circ}-70^{\circ}-\beta\). Substituting \(\beta = 60^{\circ}\), we get \(\alpha=180^{\circ}-70^{\circ}-60^{\circ}=50^{\circ}\)
Now, using the Law of Sines again to find \(b\): \(\frac{\sin50^{\circ}}{b}=\frac{\sin70^{\circ}}{5}\)

$$ b=\frac{5\sin50^{\circ}}{\sin70^{\circ}} $$

$$ b=\frac{5\times0.7660}{0.9397}=\frac{3.83}{0.9397}\approx4.1\space m $$

Answer:

\(\alpha = 60^{\circ},\beta = 50^{\circ},b = 4.1\space m\)