BC=19
Explanation
Step 1
ABE
triangle ABE is rigth triangle, then let
[tex]\begin{gathered} Angle=60 \\ adjacentside=BE \\ opposit\text{ side(the one in front of the angle)= AB=}\frac{19\sqrt[]{6}}{4} \end{gathered}[/tex]
so, we need a function that relates, angle, adjancent side and opposite side
[tex]\tan \theta=\frac{opposite\text{ side}}{\text{adjacent side}}[/tex]
replace
[tex]\begin{gathered} \tan \theta=\frac{opposite\text{ side}}{\text{adjacent side}} \\ \tan 60=\frac{AB}{\text{BE}} \\ \text{cross multiply} \\ \text{BE}\cdot\tan \text{ 60=AB} \\ \text{divide both sides by tan 60} \\ \frac{\text{BE}\cdot\tan\text{ 60}}{\tan\text{ 60}}=\frac{\text{AB}}{\tan\text{ 60}} \\ BE=\frac{\text{AB}}{\tan\text{ 60}} \\ \text{if AB=}\frac{19\sqrt[]{6}}{4} \\ BE=\frac{\frac{19\sqrt[]{6}}{4}}{\sqrt[]{3}} \\ BE=\frac{19\sqrt[]{6}}{4\sqrt[]{3}} \end{gathered}[/tex]
Step 2
BED
again, we have a rigth triangle,then let
[tex]\begin{gathered} \text{Hypotenuse}=BD \\ \text{adjacent side= BE=6.71} \\ \text{angle}=\text{ 45} \end{gathered}[/tex]
so, we need a function that relates; angle, hypotenuse and adjacent side
[tex]\cos \theta=\frac{adjacent\text{ side}}{\text{hypotenuse}}[/tex]
replace.
[tex]\begin{gathered} \cos \theta=\frac{adjacent\text{ side}}{\text{hypotenuse}} \\ \cos 45=\frac{6.71}{\text{BD}} \\ BD=\frac{6.71}{\cos \text{ 45}} \\ BD=\frac{\frac{19\sqrt[]{6}}{4\sqrt[]{3}}}{\frac{\sqrt[]{2}}{2}} \\ BD=\frac{38\sqrt[]{6}}{4\sqrt[]{6}} \\ BD=\frac{38}{4} \end{gathered}[/tex]
Step 3
finally BDE
let
angle=30
opposite side= BD
use sin function
[tex]\begin{gathered} \sin \theta=\frac{opposite\text{ side}}{\text{hypotenuse}} \\ \text{replace} \\ \sin \text{ 30=}\frac{BD}{BC} \\ BC\cdot\sin 30=BD \\ BC=\frac{BD}{\sin \text{ 30}} \\ BC=\frac{\frac{38}{4}}{\frac{1}{2}} \\ BC=\frac{76}{4}=19 \\ BC=19 \end{gathered}[/tex]
so, the answer is 19
I hop
