Answer:
56
Step-by-step explanation:
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:
It has no solution
Step-by-step explanation:
Here we have to draw the graph of y = 0.5x^2 + 3 and y = -4x^2 + 24x - 35.
I have attached the graph of these two functions.
Since the graphs of the functions didn't cut each other.
Therefore, answer is "it has no solution."
Hope this will helpful.
Thank you.
Answer:
First notice that θ is in quadrant I, where the sine and tangent will also be positive.
Then draw the angle in quadrant I. Drop a perpendicular to the x axis.
Since cosθ= 7/25, label the horizontal leg 7, and the hypotenuse 25; the vertical leg will be 24 by Pythagorean theorem.
Then sinθ=vertical/hypotenuse= 24/25 and tanθ= vertical/horizontal= 24/7
Step-by-step explanation:
C im pretty sure thats it