Question

In triunghiul dreptunghic ABC, m(<A)=90 grade, m(<B)=30 grade, AB=18 radical din 3, se duce inaltimea AD perpendicular pe BC, D apartine (BC). Calculati AC, BC si inaltimea AD

Answer 1

AC=1/2*BC⇒BC=2AC
AB²=BC²-AC²
(18√3)²=4AC²-AC²
324*3=3AC²
AC²=324
AC=18 cm
BC=2*AC=36cm
AD²=AB²-BD²
AD²=AC²-(BC-BD)²
324*3-BD²=324-(36-BD)²  ⇒972-BD²=324-1296+72BD-BD²
972-324+1296=72BD
1944=72BD ⇒BD=27
AD²=AB²-BD² ⇒AD²=972-729=243
AD=√243=9√3 cm

Answer 2

m(∡A)=90° si m(∡B)=30°⇒m(∡C)=60°
Aplicam sinus de 60 de grade si aflam ipotenuza BC
$sin60^0= \frac{Cateta\text{ } opusa}{Ipotenuza} = \frac{AB}{BC} \\ sin60^0= \frac{ \sqrt{3} }{2} = \frac{18 \sqrt{3} }{BC} \to BC= \frac{2* 18\not\sqrt{3} }{ \not\sqrt{3} } =2*18=36cm$

Δ$ABC$
$m($∡$A)=90^0 \to AB^2+AC^2=BC^2$⇔$(18 \sqrt{3} )^2+AC^2=36^2 \\ 972+AC^2=1296 \to AC= \sqrt{1296-972} = \sqrt{324} =18cm$
$AD=h= \frac{AB*AC}{BC} = \frac{18 \sqrt{3}*18 }{36} = \frac{324 \sqrt{3} }{36} =9 \sqrt{3} cm$