BC=19
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}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20Angle%3D60%20%5C%5C%20adjacentside%3DBE%20%5C%5C%20opposit%5Ctext%7B%20side%28the%20one%20in%20front%20of%20the%20angle%29%3D%20AB%3D%7D%5Cfrac%7B19%5Csqrt%5B%5D%7B6%7D%7D%7B4%7D%20%5Cend%7Bgathered%7D)
so, we need a function that relates, angle, adjancent side and opposite 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}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ctan%20%5Ctheta%3D%5Cfrac%7Bopposite%5Ctext%7B%20side%7D%7D%7B%5Ctext%7Badjacent%20side%7D%7D%20%5C%5C%20%5Ctan%2060%3D%5Cfrac%7BAB%7D%7B%5Ctext%7BBE%7D%7D%20%5C%5C%20%5Ctext%7Bcross%20multiply%7D%20%5C%5C%20%5Ctext%7BBE%7D%5Ccdot%5Ctan%20%5Ctext%7B%2060%3DAB%7D%20%5C%5C%20%5Ctext%7Bdivide%20both%20sides%20by%20tan%2060%7D%20%5C%5C%20%5Cfrac%7B%5Ctext%7BBE%7D%5Ccdot%5Ctan%5Ctext%7B%2060%7D%7D%7B%5Ctan%5Ctext%7B%2060%7D%7D%3D%5Cfrac%7B%5Ctext%7BAB%7D%7D%7B%5Ctan%5Ctext%7B%2060%7D%7D%20%5C%5C%20BE%3D%5Cfrac%7B%5Ctext%7BAB%7D%7D%7B%5Ctan%5Ctext%7B%2060%7D%7D%20%5C%5C%20%5Ctext%7Bif%20AB%3D%7D%5Cfrac%7B19%5Csqrt%5B%5D%7B6%7D%7D%7B4%7D%20%5C%5C%20BE%3D%5Cfrac%7B%5Cfrac%7B19%5Csqrt%5B%5D%7B6%7D%7D%7B4%7D%7D%7B%5Csqrt%5B%5D%7B3%7D%7D%20%5C%5C%20BE%3D%5Cfrac%7B19%5Csqrt%5B%5D%7B6%7D%7D%7B4%5Csqrt%5B%5D%7B3%7D%7D%20%5Cend%7Bgathered%7D)
Step 2
BED
again, we have a rigth triangle,then let
so, we need a function that relates; angle, hypotenuse and adjacent side
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=\frac{38}{4} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ccos%20%5Ctheta%3D%5Cfrac%7Badjacent%5Ctext%7B%20side%7D%7D%7B%5Ctext%7Bhypotenuse%7D%7D%20%5C%5C%20%5Ccos%2045%3D%5Cfrac%7B6.71%7D%7B%5Ctext%7BBD%7D%7D%20%5C%5C%20BD%3D%5Cfrac%7B6.71%7D%7B%5Ccos%20%5Ctext%7B%2045%7D%7D%20%5C%5C%20BD%3D%5Cfrac%7B%5Cfrac%7B19%5Csqrt%5B%5D%7B6%7D%7D%7B4%5Csqrt%5B%5D%7B3%7D%7D%7D%7B%5Cfrac%7B%5Csqrt%5B%5D%7B2%7D%7D%7B2%7D%7D%20%5C%5C%20BD%3D%5Cfrac%7B38%5Csqrt%5B%5D%7B6%7D%7D%7B4%5Csqrt%5B%5D%7B6%7D%7D%20%5C%5C%20BD%3D%5Cfrac%7B38%7D%7B4%7D%20%5Cend%7Bgathered%7D)
Step 3
finally BDE
let
angle=30
opposite side= BD
use sin function
so, the answer is 19
I hop
Answer:
D
Step-by-step explanation:
1. Remind that
Here you can use trigonometric table to find the value of 
, or you can remind that 
and 
because special right triangle 45°-45°-90° is isosceles (with congruent oppposite and adjacent legs) and the ratio between the opposite leg and adjacent leg is equal to 1.
2. Now
because
True. Two lines intersect in a single point.
Answer: 2.5 ounces/ pint
You can firstly turn your fractions into decimals 5/8 is equal to .625 and 1/4 is .25
Then you can divide .625 by .25 which is 2.5, So the unit rate is 2.5 ounces per pint.
