anonymous
  • anonymous
Find the measure of Angle e?? need major help
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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welshfella
  • welshfella
First count the number of sides in the polygon How many?
anonymous
  • anonymous
9

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anonymous
  • anonymous
9-2=7 7 times 180 right
welshfella
  • welshfella
right now there a formula for finding the total number of angles in a polygon form its number of sides. if i recall correctly its 180(n - 2) i'll check that out.
anonymous
  • anonymous
1,260 is the sum
welshfella
  • welshfella
yes thats right so you now set up an equation
anonymous
  • anonymous
what would it look like?
welshfella
  • welshfella
6x - 1 + 7x + 6 etc = 1260 solve this for x and angle e = 7x + 9
anonymous
  • anonymous
170 is the answer thanks!
anonymous
  • anonymous
can i ask you another question??
welshfella
  • welshfella
ok
anonymous
  • anonymous
find the value of c. a. 10 b.5 c.square root 2 d. 10 square root 2
anonymous
  • anonymous
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welshfella
  • welshfella
this right angled triangle is also isosceles because the unmarked angle is also 45 degrees. So the other leg is also 5 sqrt2 you use the pythagoras theorem to solve this c^2 = (5 sqrt2)^2 + (5 sqrt2)^2
welshfella
  • welshfella
(5 sqrt2)^2 = 5^2 * 2 = ?
anonymous
  • anonymous
10??
welshfella
  • welshfella
the final answer?
anonymous
  • anonymous
c^2 = (5 sqrt2)^2 + (5 sqrt2)^2, does that equal 10?
welshfella
  • welshfella
no (5 sqrt2)^2 = 5^2 * 2
welshfella
  • welshfella
= 25 8 2 = 50
welshfella
  • welshfella
* thats * not 8
anonymous
  • anonymous
what do you do with the 50 now?
welshfella
  • welshfella
c^2 = 50 + 50
welshfella
  • welshfella
i've assumed you are familiar with the pythagoras theorem
anonymous
  • anonymous
which is 100, and the final answer is c, so would it be 10?
welshfella
  • welshfella
yep
anonymous
  • anonymous
yes a^2 +b^2 = C^2
anonymous
  • anonymous
i have another problem like this with a scalene triangle
welshfella
  • welshfella
ok gtg now hope this has helped
anonymous
  • anonymous
thank you

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