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The Solution:
Represent the problem in a diagram:
Required:
To find the values of:
[tex]\begin{gathered} \angle A \\ \angle B \\ c \end{gathered}[/tex]Step 1:
Use the Law of cosines to find the value of c.
[tex]\begin{gathered} c^2=a^2+b^2-2ab\cos C \\ \text{ where} \\ c=? \\ a=5.10m \\ b=8.73m \\ C=108.5^o \end{gathered}[/tex]Substituting these values, we get
[tex]\begin{gathered} c^2=5.10^2+8.73^2-(2\times5.10\times8.73\times\cos108.5) \\ \\ c^2=130.47761 \\ \\ c=\sqrt{130.47761}=11.4227\approx11.42m \end{gathered}[/tex]Step 2:
Use the Law of sines to find angle A.
[tex]\frac{\sin A}{a}=\frac{\sin C}{c}[/tex]Substituting, we get:
[tex]\begin{gathered} \frac{\sin A}{5.10}=\frac{\sin108.5}{11.42} \\ So, \\ \\ \sin A=\frac{5.10\times\sin108.5}{11.42}=0.423507 \\ \\ A=\sin^{-1}(0.423507)=25.0562\approx25.1^o \end{gathered}[/tex]Step 3:
Find angle B.
By the sum of angles in a triangle:
[tex]\begin{gathered} \angle A+\angle B+\angle C=180^o \\ 25.1+\angle B+108.5=180 \\ \angle B=180-(108.5+25.1) \\ \angle B=180-133.6=46.4^o \end{gathered}[/tex]Therefore, the correct answers are:
[tex]\begin{gathered} \angle A=25.1^o \\ \\ \angle B=46.4^o \\ \\ c=11.42m \end{gathered}[/tex]