Question

Concave polygon is such that of its internal measures either 60° or 300° if this polygon contains twenty 300° angles then the number of 60° angles in it is

Answer 1

Answer:

The number of 60° angles in given conclave polygon is 23

Step-by-step explanation:

Given -> Any conclave polygon having "n" sides. Its internal angles are 60° and 300° only.

->There are 20 angles of 300° each.

FORMULA USED->i) Sum of Interior angles of polygon =(n-2)×180°

                                 where n = number of sides of polygon

ii) number of internal angles in a polygon is equal to no of sides of polygon.

SOLUTION->  In given polygon no of sides is 'n' hence no of internal angles is also 'n'.

->sum of internal angles = (sum of given twenty 300° angles)+(sum of remaining 60° angles.

->(n-2)×180°=(20×300°) + { (n-20)×60°}

->180°n-360°=6000°+ 60°n-1200°

->180°n-60°n=6000°-1200+360°

->120°n=5160°

->n=$\frac{5160}{120}$

->n=43

Hence total no. of angles is equal to no. of sides i.e. 43.

so remaining no. of 60° angles = n-20=43-20= 23

Answer 2

Step-by-step explanation:

The number of 60° angles in given conclave polygon is 23

Step-by-step explanation:

Given -> Any conclave polygon having "n" sides. Its internal angles are 60° and 300° only.

->There are 20 angles of 300° each.

FORMULA USED->i) Sum of Interior angles of polygon =(n-2)×180°

where n = number of sides of polygon

ii) number of internal angles in a polygon is equal to no of sides of polygon.

SOLUTION-> In given polygon no of sides is 'n' hence no of internal angles is also 'n'.

->sum of internal angles = (sum of given twenty 300° angles)+(sum of remaining 60° angles.

->(n-2)×180°=(20×300°) + { (n-20)×60°}

->180°n-360°=6000°+ 60°n-1200°

->180°n-60°n=6000°-1200+360°

->120°n=5160°

->n=\frac{5160}{120}

120

5160

->n=43

Hence total no. of angles is equal to no. of sides i.e. 43.

so remaining no. of 60° angles = n-20=43-20= 23