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We have the following triangle
We have the hypotenuse and all the angles so we can use the law of sines
[tex]\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}[/tex]In this case take us a = 4, b = QP andc = RQ
First, we solve QP
[tex]\begin{gathered} \frac{4}{\sin(90)}=\frac{b}{\sin(30)} \\ b=\frac{4}{\sin(90)}\cdot\sin (30) \\ b=2 \end{gathered}[/tex]Second. we solve RQ
[tex]\begin{gathered} \frac{4}{\sin(90)}=\frac{c}{\sin(60)} \\ c=\frac{4}{\sin(90)}\cdot\sin (60) \\ c=2\sqrt[]{3} \end{gathered}[/tex]These are the solutions
To check this we can take out the hypotenuse using the Pythagoras theorem and check that
[tex]\begin{gathered} H=\sqrt[]{(2)^2+(2\sqrt[]{3})^2} \\ H=\sqrt[]{4+(4\cdot3)} \\ H=\sqrt[]{4+12} \\ H=\sqrt[]{16} \\ H=4 \end{gathered}[/tex]This is correct
In conclusion, these answers are:
[tex]\begin{gathered} QP=2 \\ RQ=2\sqrt[]{3} \end{gathered}[/tex]Thanks for taking the time to read Help finding the missing side lengths for RQ and QP. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!
Rewritten by : Barada
