Check the picture below to the left, let's use those sides with the law of sines
![\textit{Law of sines} \\\\ \cfrac{sin(\measuredangle A)}{a}=\cfrac{sin(\measuredangle B)}{b}=\cfrac{sin(\measuredangle C)}{c} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{sin(14^o)}{97}=\cfrac{sin(84^o)}{XZ}\implies XZ = \cfrac{97\cdot sin(84^o)}{sin(14^o)}\implies XZ \approx 398.76 \\\\\\ \stackrel{\textit{now using SOH CAH TOA}}{cos(82^o) = \cfrac{XW}{XZ}}\implies XZcos(82^o)=XW \\\\\\ 398.76cos(82^o)\approx XW\implies 55.497\approx XW\implies \stackrel{\textit{rounded up}}{55=XW}](https://tex.z-dn.net/?f=%5Ctextit%7BLaw%20of%20sines%7D%20%5C%5C%5C%5C%20%5Ccfrac%7Bsin%28%5Cmeasuredangle%20A%29%7D%7Ba%7D%3D%5Ccfrac%7Bsin%28%5Cmeasuredangle%20B%29%7D%7Bb%7D%3D%5Ccfrac%7Bsin%28%5Cmeasuredangle%20C%29%7D%7Bc%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7Bsin%2814%5Eo%29%7D%7B97%7D%3D%5Ccfrac%7Bsin%2884%5Eo%29%7D%7BXZ%7D%5Cimplies%20XZ%20%3D%20%5Ccfrac%7B97%5Ccdot%20sin%2884%5Eo%29%7D%7Bsin%2814%5Eo%29%7D%5Cimplies%20XZ%20%5Capprox%20398.76%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bnow%20using%20SOH%20CAH%20TOA%7D%7D%7Bcos%2882%5Eo%29%20%3D%20%5Ccfrac%7BXW%7D%7BXZ%7D%7D%5Cimplies%20XZcos%2882%5Eo%29%3DXW%20%5C%5C%5C%5C%5C%5C%20398.76cos%2882%5Eo%29%5Capprox%20XW%5Cimplies%2055.497%5Capprox%20XW%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Brounded%20up%7D%7D%7B55%3DXW%7D)
I could be wrong but I think it’s 144,000m for the area
Well 120 is 1/3 of the square so the 9ft is 2/3 of the square so it is B
Answer:
20
Step-by-step explanation:
→ First find the value of x and y
x + y = 5
x - y = 4
→ Add the equations to cancel out the y's
2x = 9
→ Divide both sides by 2 to find the value of x
x = 4.5
→ Substitute x = 4.5 back into x - y = 4 to find the value of y
4.5 - y = 4
→ Minus 4.5 from both sides to isolate -y
-y = -0.5
→ Multiply everything by -1
y = 0.5
→ Substitute x = 4.5 and y = 0.5 into x² - y²
4.5² - 0.5² = 20.25 - 0.25 = 20
