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]
