Answer:
[tex]First\ angle = x = 40^\circ\\\\Second\ angle = 2x = 2 \times 40 = 80^\circ\\\\Third \ angle = 3x = 3 \times 40 = 120^\circ\\\\Fourth \ angle = 4x = 4\times 40 = 160^\circ\\\\Fifth\ angle = (4x - 20) = ( 4 \times 40) - 20 = 160 -20 = 140^\circ[/tex]
Step-by-step explanation:
Let the first angle be = x
Given:
Second angle is twice first = 2x
Third angle is three times first = 3x
Also given:
Second angle is half of fourth angle, that is:
[tex]2x = \frac{1}{2} \times 4 ^{th} angle\\\\4^{th} angle = 2x \times 2 = 4x[/tex]
Fifth angle is 20° less than fourth angle, that is:
[tex]4x - 20^\circ[/tex]
Sum of interior angles of a polygon with n sides = ( n - 2 ) x 180°
Here n = 5 ,
therefore sum of interior angles = ( 5 - 2 ) x 180 = 3 x 180 = 540°.
That is ,
x + 2x + 3x + 4x + ( 4x - 20 ) = 540
14x - 20 = 540
14x = 540 + 20
14x = 560
x = 40
[tex]Therefore, \\\\First\ angle = x = 40^\circ\\\\Second\ angle = 2x = 2 \times 40 = 80^\circ\\\\Third \ angle = 3x = 3 \times 40 = 120^\circ\\\\Fourth \ angle = 4x = 4\times 40 = 160^\circ\\\\Fifth\ angle = (4x - 20) = ( 4 \times 40) - 20 = 160 -20 = 140^\circ[/tex]
