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
It can be used in statics for the major of Mechanical Engineering. You can find the total amount of weight on beams by forces.
Let m be mean
Mean= sum/ n
Mean= (1720+1687+1367+1614+1460+1867+1436) / 7
m= 11151 / 7
M= 1593
Mean= 1593
Standard deviation
|x-m|^2
For 1st: |1720-1593|^2=8836
For 2nd: |1687-1593|^2=10201
For 3rd: |1367-1593|^2=51076
For 4th: |1614-1593|^2=441
For 5th: |1460-1593|^2=1689
For 6th: |1867-1593|^2=75076
For 7th: |1436-1593|^2=24649
Summation of |x-m|^2 = 171968
Standard deviation sample formula is:
S.D = sqrt((summation of |x-m|^2) / n-1)
S.D=sqrt(171968/6)
S.D=sqrt(28661.33)
S.D=169.30
Standard deviation is 169.30
Answer: 65/100 or simplified
Step-by-step explanation:
