Answer:
Part a) [tex]BC=10\ units[/tex]
Part b) [tex]AC=13.66\ units[/tex]
Step-by-step explanation:
step 1
Find the length side BC
Applying the law of sines
we know that
[tex]\frac{AB}{sin(C)}=\frac{BC}{sin(A)}[/tex]
we have
[tex]AB=5\sqrt{2}\ units[/tex]
[tex]A=45^o[/tex]
[tex]C=30^o[/tex]
substitute
[tex]\frac{5\sqrt{2}}{sin(30^o)}=\frac{BC}{sin(45^o)}[/tex]
solve for BC
[tex]BC=\frac{5\sqrt{2}}{sin(30^o)}(sin(45^o))[/tex]
[tex]BC=10\ units[/tex]
step 2
Find the measure of angle B
we know that
The sum of the interior angles in a triangle must be equal to 180 degrees
so
[tex]m\angle A+m\angle B+m\angle C=180^o[/tex]
substitute the given values
[tex]45^o+m\angle B+30^o=180^o[/tex]
[tex]75^o+m\angle B=180^o[/tex]
[tex]m\angle B=180^o-75^o[/tex]
[tex]m\angle B=105^o[/tex]
step 3
Find the length side AC
Applying the law of sines
we know that
[tex]\frac{AB}{sin(C)}=\frac{AC}{sin(B)}[/tex]
we have
[tex]AB=5\sqrt{2}\ units[/tex]
[tex]A=45^o[/tex]
[tex]B=105^o[/tex]
substitute
[tex]\frac{5\sqrt{2}}{sin(30^o)}=\frac{AC}{sin(105^o)}[/tex]
solve for AC
[tex]AC=\frac{5\sqrt{2}}{sin(30^o)}(sin(105^o))[/tex]
[tex]AC=13.66\ units[/tex]
