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

Answer:

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

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:

                                                                   $2x = \frac{1}{2} \times 4 ^{th} angle\\\\4^{th} angle = 2x \times 2 = 4x$

 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

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