You can use the sine rule to solve this question,
[tex]\frac{|NO|}{\sin(45)} =\frac{6}{\sin30}[/tex]
This implies that
[tex]|NO|} =\frac{6}{\sin30} \times \sin(45)[/tex]
[tex]|NO|} =6 \sqrt(2)=8.49[/tex]
[tex]<MNO+45+30=180[/tex]
[tex]<MNO=180-45-30[/tex]
[tex]<MNO=105[/tex]
Using the sine rule again
[tex]\frac{|MO|}{\sin(105)} =\frac{6}{\sin30}[/tex]
This implies that
[tex]|MO|} =\frac{6}{\sin30} \times \sin(105)[/tex]
[tex]|MO|} =11.59[/tex]
To find the length of the sides NO and MO, we can use sine law.
We have been given that:
[tex]m\angle M=45^{\circ}, m\angle O=30^{\circ}[/tex]
We know that the sum of 3 angles of a triangle make [tex]180^\circ[/tex].
So,
[tex]\angle M+\angle N+\angle O=180^\circ[/tex]
Substituting the values of the given angles we get,
[tex]45^\circ+30^\circ+ \angle N =180^\circ[/tex]
[tex]\angle N=180^\circ -45^\circ-30^\circ=105^\circ[/tex]
Now,
Let the length of the side NO be 'x' units and the length of the side MO be 'y' units.
Using the sine law we can state that (refer the attached figure):
[tex]\frac{sin M}{x}=\frac{sin N}{y}=\frac{sin O}{6}[/tex]
Putting the values we get:
[tex]\frac{sin 45^{\circ}}{x}=\frac{sin 105^\circ}{y}=\frac{sin 30^\circ}{6}[/tex]
Now, using
[tex]\frac{sin 45^{\circ}}{x}=\frac{sin 30^\circ}{6}[/tex]
Putting the values we get,
[tex]\frac{ 0.707}{x}=\frac{0.5}{6}[/tex]
Therefore,
[tex]x=0.707 \times \frac{6}{0.5}=8.48 \approx 8.5[/tex]
Therefore, the length of the side NO is 8.5 units
Using the same equation, we can find the length of the side MO,
[tex]\frac{sin 105^\circ}{y}=\frac{sin 30^\circ}{6}[/tex]
Putting the values of the angles we get,
[tex]\frac{0.965}{y}=\frac{0.5}{6}[/tex]
So,
[tex]y=0.965 \times \frac{6}{0.5}=11.58 \approx 11.6[/tex]
Therefore, the length of the side MO is 11.6 units.
