Question

if the exterior angles of a polygon are x°,(x+5)°,(x+15)°,(x+10)°and (x+20)°,find the value of x​

Answer 1

ANSWER:

Given:

• Exterior Angles of a polygon = x°, (x+5)°, (x+15)°, (x+10)° and (x+20)°.

To Find:

• Value of x

Solution:

We know that,

⇒ Sum of all exterior angles of a polygon = 360°

So,

⇒ x° + (x + 5)° + (x + 10)° + (x + 15)° + (x + 20)° = 360°

Opening the brackets,

⇒ x° + x° + 5° + x° + 10° + x° + 15° + x° + 20° = 360°

Adding all the x, and the numbers,

⇒ 5x° + 50° = 360°

Transposing 50° to RHS,

⇒ 5x° = 360° - 50°

⇒ 5x° = 310°

Transposing 5 to RHS,

⇒ x° = (310/5) °

⇒ x = 62

So, the value of x is 62°.

Formula Used:

• Sum of all exterior angles of a polygon = 360°

Answer 2

$\large {\maltese {\textsf {\textbf {\underline { Given :}}}}}$

• Exterior angles of polygon are x° , (x + 5)° , (x + 15)° , ( x + 10)° & ( x + 20)°

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$\large {\maltese {\textsf {\textbf {\underline { To \; find :}}}}}$

• Value of x

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$\large {\maltese {\textsf {\textbf {\underline { Solution :}}}}}$

As, we know that :-

• Sum of exterior angles of a polygon is 360°

So,

$\implies \sf { x° + (x + 5)° + (x + 15)° + ( x + 10)° + ( x + 20)° = 360° }$

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$\implies \sf { 5x + 50 = 360 }$

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$\implies \sf { 5x = 360 - 50}$

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$\implies \sf { 5x = 310}$

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$\implies \sf { x = \dfrac {310}{5}}$

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$\implies \sf { x = 62}$

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$\sf {\underline { Therefore, \; value \; of \; x = 62°}}$