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To solve the exercise you can use the law of cosine, which applies to any triangle:
[tex]b^2=a^2+c^2-2ac\cdot\cos (B)[/tex]Where
So, in this case, you have
[tex]\begin{gathered} p^2=o^2+q^2-2oq\cdot\cos (P) \\ p^2=(3.8cm)^2+(1.7cm)^2-2(3.8cm)(1.7cm)\cdot\cos (96\text{\degree)} \\ p^2=14.44cm^2+2.89cm^2-12.92cm^2\cdot\cos (96\text{\degree)} \\ p^2=17.33\operatorname{cm}^2-(-1.35cm^2) \\ p^2=17.33\operatorname{cm}+1.35cm^2 \\ p^2=18.68\operatorname{cm}^2 \\ \text{ Apply square root to both sides of the equation} \\ \sqrt[]{p^2}=\sqrt[]{18.68\operatorname{cm}^2} \\ p=4.32\operatorname{cm} \end{gathered}[/tex]Now, you can use the law of sine, which applies to any triangle:
[tex]\frac{a}{\sin (A)}=\frac{b}{\sin (B)}=\frac{c}{\sin (C)}[/tex]In this case, you have
[tex]\begin{gathered} \frac{q}{\sin(Q)}=\frac{p}{\sin(P)} \\ \frac{1.7\operatorname{cm}}{\sin(Q)}=\frac{4.32\operatorname{cm}}{\sin(96\text{\degree})} \\ \text{ Apply cross product} \\ 1.7\operatorname{cm}\cdot\sin (96\text{\degree})=\sin (Q)\cdot4.32\operatorname{cm} \\ \text{ Divide by 4.32 cm from both sides of the equation} \\ \frac{1.7\operatorname{cm}\cdot\sin(96\text{\degree})}{4.32\operatorname{cm}}=\frac{\sin(Q)\cdot4.32\operatorname{cm}}{4.32\operatorname{cm}} \\ \frac{1.7\operatorname{cm}\cdot\sin(96\text{\degree})}{4.32\operatorname{cm}}=\sin (Q) \\ 0.39=\sin (Q) \\ \text{ Apply the inverse function }\sin ^{-1}(\theta)\text{ both sides of the equation} \\ \sin ^{-1}(0.39)=\sin ^{-1}(\sin (Q)) \\ 23.0\text{\degree}=Q \end{gathered}[/tex]Therefore, the measure of angle Q is 23 degrees.
