Question

I came up the answer as 57. I will attach my note, can you check?

Answer 1

BC=19

Step-by-step explanation

Step 1

ABE

triangle ABE is rigth triangle, then let

`\begin{gathered} Angle=60 \\ adjacentside=BE \\ opposit\text{ side(the one in front of the angle)= AB=}\frac{19\sqrt[]{6}}{4} \end{gathered}`

so, we need a function that relates, angle, adjancent side and opposite side

`\tan \theta=\frac{opposite\text{ side}}{\text{adjacent side}}`

replace

`\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}`

Step 2

BED

again, we have a rigth triangle,then let

`\begin{gathered} \text{Hypotenuse}=BD \\ \text{adjacent side= BE=6.71} \\ \text{angle}=\text{ 45} \end{gathered}`

so, we need a function that relates; angle, hypotenuse and adjacent side

`\cos \theta=\frac{adjacent\text{ side}}{\text{hypotenuse}}`

replace.

`\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=(38)/(4) \end{gathered}`

Step 3

finally BDE

let

angle=30

opposite side= BD

use sin function

`\begin{gathered} \sin \theta=\frac{opposite\text{ side}}{\text{hypotenuse}} \\ \text{replace} \\ \sin \text{ 30=}(BD)/(BC) \\ BC\cdot\sin 30=BD \\ BC=\frac{BD}{\sin \text{ 30}} \\ BC=((38)/(4))/((1)/(2)) \\ BC=(76)/(4)=19 \\ BC=19 \end{gathered}`

so, the answer is 19

I hop