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.

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05-07-2021
Mathematics

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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.

Sagot :

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Sreedevi102idn Sreedevi102idn · Answer 1 · 2021-07-06T03:21:03-04:00

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]

Sagot :

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