Question

PLZZZZ HELP ASAP !!!!!

Answer 1

Answer:

Step-by-step explanation:

a.

$AC=\sqrt{AB^2-BC^2} =\sqrt{13^2-5^2} =\sqrt{169-25} =\sqrt{144} =12$

applying sine formula

AB=c

AC=b

BC=a

$\frac{a}{sin A} =\frac{b}{sin B} =\frac{c}{sin C} \\\\\frac{5}{sin A} =\frac{12}{sin B} =\frac{13}{sin 90} \\using~last~two\\\frac{12}{sin B}= \frac{13}{1} \\sin B=\frac{12}{13}\\ B=sin^{-1}(\frac{12}{13}) \approx 67.38 ^\circ$

b.

third angle ∠Y=180-(30+90)=180-120=60

using sine formula

$\frac{xz}{sin Y} =\frac{xy}{sin Z} = \frac{yz}{sin X} \\\frac{30}{sin 60} =\frac{xy}{sin 90} =\frac{h}{sin 30} \\using ~first~and~last\\\frac{30}{\frac{\sqrt{3} }{2} } =\frac{h}{\frac{1}{2} } \\2h=\frac{60}{\sqrt{3} } \\h=\frac{30}{\sqrt{3} } \times \frac{\sqrt{3} }{\sqrt{3} } =10\sqrt{3}$