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)
Answer:
area = 693
Step-by-step explanation:
The area of a quadrilateral can be found using the formula; base * height
It is given that the height is 33, the base is 21;
Multiply the two
21 *33 = 693
Answer:
Part 1) m∠EFG=94°
Part 2) m∠GFH=86°
Step-by-step explanation:
we know that
m∠EFG+m∠GFH=180° -----> by linear pair (given problem)
we have
m∠EFG=3n+22
m∠GFH=2n+38
substitute the values
(3n+22)°+(2n+38)°=180°
Solve for n
(5n+60)=180
5n=180-60
5n=120
n=24
<em>Find the measure of angle EFG</em>
m∠EFG=3n+22
substitute the value of n
m∠EFG=3(24)+22=94°
<em>Find the measure of angle GFH</em>
m∠GFH=2n+38
substitute the value of n
m∠GFH=2(24)+38=86°
