Supplementary angles , when added, = 180
complimentary angles, when added, = 90
< AQC + < GQC = 180.....supplementary
< BQD + < DQE = 90.......complimentary
< CQE + < EQF = 90.......complimentary
< GQF , < FQE.....neither
< BQC + < DQC = 90....complimentary
< W and < X are supplementary...
if < W = 37, then < X = (180 - 37) = 143
< S and < T are complimentary
if < S = 64, then < T = (90 - 64) = 26
< C and < D are supplementary
if < C = 83, the < D = (180 - 83) = 97
cant read all of the last one.....but if they are complimentary, and
< U = 41, then the other angle is : (90 - 41) = 49
First the equation is 3+4, which equals 7
Answer:
302.02 units²
Step-by-step explanation:
To do this, we can first solve for the height of the trapezoid. We can do this using sin:
sin 70= 
14· sin 70≈13.16
Now, we can solve for the base of the small triangle on the right by:
cos 70= 
14 · cos 70 ≈4.79
Now, find the area of the triangle using
:
V= (4.79)(13.16)/2≈30.86
Now, we can find the area of the rectangle in the middle by:
15· 13.16 (the height)= 197.4
Now, we need to solve for the triangle on the left. We will need to first find the base by subtracting the bases of the other figures from 31:
31-4.79-15= 11.21
Now, find the area of the triangle:
V=(11.21)(13.16)/2= 73.76
Add all of the areas, giving you:
73.76+30.86+197.4= 302.02 units²
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Step-by-step explanation: