Answer:
Step-by-step explanation:
a.
[tex]AC=\sqrt{AB^2-BC^2} =\sqrt{13^2-5^2} =\sqrt{169-25} =\sqrt{144} =12\\[/tex]
applying sine formula
AB=c
AC=b
BC=a
[tex]\frac{a}{sin A} =\frac{b}{sin B} =\frac{c}{sin C} \\\\\frac{5}{sin A} =\frac{12}{sin B} =\frac{13}{sin 90} \\using~last~two\\\frac{12}{sin B}= \frac{13}{1} \\sin B=\frac{12}{13}\\ B=sin^{-1}(\frac{12}{13}) \approx 67.38 ^\circ[/tex]
b.
third angle ∠Y=180-(30+90)=180-120=60
using sine formula
[tex]\frac{xz}{sin Y} =\frac{xy}{sin Z} = \frac{yz}{sin X} \\\frac{30}{sin 60} =\frac{xy}{sin 90} =\frac{h}{sin 30} \\using ~first~and~last\\\frac{30}{\frac{\sqrt{3} }{2} } =\frac{h}{\frac{1}{2} } \\2h=\frac{60}{\sqrt{3} } \\h=\frac{30}{\sqrt{3} } \times \frac{\sqrt{3} }{\sqrt{3} } =10\sqrt{3}[/tex]
