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The dimension of the triangle PQR given in the question is as shown below:
[tex]\begin{gathered} RQ=\text{?},m\angle RQP=90^0 \\ QP=\text{?},m\angle QPR=60^0 \\ RP=4,m\angle PRQ=30^0 \end{gathered}[/tex]Using trigonometric ratio of sine of angle PRQ,
[tex]\begin{gathered} \sin =\frac{opposite}{\text{hypothenuse}} \\ \sin 30^0=\frac{QP}{4} \\ QP=4\times\sin 30^0 \\ QP=4\times0.5000 \\ QP=2 \end{gathered}[/tex]Since two sides are known, sides RP and QP, we can derive RQ using the Pythaogoras theorem as shown below:
[tex]\begin{gathered} RP^2=QP^2+RQ^2 \\ 4^2=2^2+RQ^2 \\ 16=4+RQ^2 \\ 16-4=RQ^2 \\ 12=RQ^2 \\ RQ^2=12 \\ RQ=\sqrt[]{12} \\ RQ=3.464 \end{gathered}[/tex]Hence,
RQ = 3.464 units
QP = 4 units
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Rewritten by : Barada
