Question

In a pentagon, the second angle is twice the first but half of the fourth, third angle is three times the first and the fifth angle is 20° less than the fourth. Find all the angles.​

Answer 1

`First\ angle = x = 40^\circ\\\\Second\ angle = 2x = 2 * 40 = 80^\circ\\\\Third \ angle = 3x = 3 * 40 = 120^\circ\\\\Fourth \ angle = 4x = 4* 40 = 160^\circ\\\\Fifth\ angle = (4x - 20) = ( 4 * 40) - 20 = 160 -20 = 140^\circ`

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:

`2x = (1)/(2) * 4 ^(th) angle\\\\4^(th) angle = 2x * 2 = 4x`

Fifth angle is 20° less than fourth angle, that is:

`4x - 20^\circ`

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

`Therefore, \\\\First\ angle = x = 40^\circ\\\\Second\ angle = 2x = 2 * 40 = 80^\circ\\\\Third \ angle = 3x = 3 * 40 = 120^\circ\\\\Fourth \ angle = 4x = 4* 40 = 160^\circ\\\\Fifth\ angle = (4x - 20) = ( 4 * 40) - 20 = 160 -20 = 140^\circ`