20×1=20
20×2=40
20×3=60
20×4=80
20×5=100
20×6=120
20×7=140
20×8=160
20×9=180
20×10=200
(Only include this if the table is up to 12)
20×11=220
20×12=240
Answer:
Actually it's not polygon. it's a nonagon. With r=8.65mm″, the law of cosines gives us side a:
a=√{b²+c²−2bc×cos40°}
a=√{149.645−149.645cos40°}
Area Nonagon = (9/4)a²cos40°
=9/4[149.645−149.645cos40°]cot20°
=336.70125[1−cos(40°)]cot(20°)
Applying an identity for the cos(40°) does not get us very far…
= 336.70125[1−(cos2(20°)−1)]cot(20°)
= 336.70125[2−cos2(20°)]cot(20°)
= 336.70125[2−(1−sin2(20°))]cot(20°)
= 336.70125[1+sin2(20°)]cos(20°)sin(20°)
= 336.70125[cot(20°)+sin(20°)cos(20°)]mm²
M∠LON=77 ∘ m, angle, L, O, N, equals, 77, degrees \qquad m \angle LOM = 9x + 44^\circm∠LOM=9x+44 ∘ m, angle, L, O, M, equals, 9,
timama [110]
Answer:
Step-by-step explanation:
Given
<LON = 77°
<LOM = (9x+44)°
<MON = (6x+3)°
The addition postulate is true for the given angles since tey have a common point O:
<LON = <LOM+<MON
Since we are not told what to find we can as well look for the value of x, <LOM and <MON
Substitute the given parameters and get x
77 = 9x+44+6x+3
77 = 15x+47
77-47 = 15x
30 = 15x
x = 30/15
x = 2
Get <LOM:
<LOM = 9x+44
<LOM = 9(2)+44
<LOM = 18+44
<LOM = 62°
Get <MON:
<MON = 6x+3
<MON = 6(2)+3
<MON = 12+3
<MON = 15°
