Problems from Solution of triangles

Development:

Let's identify the information of the problem using the following figure:

Let's observe that we know one side and two angles. Let's apply the sine theorem: $${\displaystyle \frac{a}{\sin (A)}}={\displaystyle \frac{b}{\sin (B)}}={\displaystyle \frac{c}{\sin (C)}}$$

We see, that it would be necessary to know the angle $A$, but this is not a problem, since $$A={180}^{\circ }-B-C={180}^{\circ }-{45}^{\circ }-{105}^{\circ }={30}^{\circ }$$ Thus: $$b={\displaystyle \frac{a\cdot \sin (B)}{\sin (A)}}={\displaystyle \frac{a\cdot \sin ({45}^{\circ })}{\sin ({30}^{\circ })}}={\displaystyle \frac{6{\displaystyle \frac{\sqrt{2}}{2}}}{{\displaystyle \frac{1}{2}}}}=6\sqrt{2}\text{m}$$

Now we already know 2 sides and 2 angles. We can then apply the sine or the cosine theorem. We are going to apply again the sine one. So we have: $$c={\displaystyle \frac{a\cdot \sin (C)}{\sin (A)}}={\displaystyle \frac{6\cdot \sin ({105}^{\circ })}{\sin ({30}^{\circ })}}={\displaystyle \frac{6{\displaystyle \frac{1}{4}}(\sqrt{6}+\sqrt{2})}{{\displaystyle \frac{1}{2}}}}=3(\sqrt{6}+\sqrt{2})\text{m}$$

Solution:

$A={30}^{\circ }$

$b=6\sqrt{2}$ m

$b=3(\sqrt{6}+\sqrt{2})$ m

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