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27 votes
27 votes
If the exterior angles of a pentagon are x° 2x° (x+30)°,(x-10)° and (x+40)° find x

a) 30
b) 50
c) 60
d) 80

User Sepehrom
by
2.9k points

2 Answers

19 votes
19 votes

Answer:


\fbox {b) 50}

Explanation:

The sum of the exterior angles of any polygon (including a pentagon) is equal to 360 degrees.

Then :

⇒ x + 2x + x + 30 + x - 10 + x + 40 = 360

⇒ 6x + 60 = 360

⇒ 6x = 300

x = 50

User VeikkoW
by
3.3k points
21 votes
21 votes

Answer:


\huge\boxed{\sf x = 50}

Explanation:

Statement:

  • The exterior angles of a pentagon add up 360 degrees.

Given angles:

  • 2x°
  • (x+30)°
  • (x-10)°
  • (x+40)°

Solution:

x + 2x + x + 30 + x - 10 + x + 40 = 360

6x + 60 = 360

Subtract 60 to both sides

6x = 360 - 60

6x = 300

Divide 6 to both sides

x = 300/6

x = 50


\rule[225]{225}{2}

User Huuuk
by
2.9k points