You have two regular polygons A & B
. The ratio of the interior angles of
A & B
is

21: 22
. The ratio of the exterior angles is
3:2
. How many sides does each polygon have?

First polygon has 16 sides

Second polygon has 24 sides

Step-by-step explanation:

Let the number of sides for first polygon be a and that for the second polygon be b

formula for sum of interior angle is;

180n-360

So for individual interior angle we have;

(180n-360)/n

So for polygon A

(180a-360)/a

For polygon B

(180b-360)/b

Let’s divide both to equate to the ratio

(180a-360/a)/(180b-360)/b) = 21/22

Factor out 180

(a-2)/a * b/b-2 = 21/22

b(a-2)/a(b-2) = 21/22

22b(a-2) = 21a(b-2) •••••• (i)

For the second part

Sum of exterior angle of any polygon is 360

Individual exterior angle is 360/n

for the first polygon;

360/a ;

For the second 360/b

Thus;

b/a = 3/2

2b = 3a ••••••••(ii)

b = 3a/2

11(2b)(a-2) = 21a(b-2)

11(3a)(a-2) = 21a( 3a/2-2)

33a(a-2) = 21a(3a-4)/2

22a(a-2) = 7a(3a-4)

22a^2 - 44a= 21a^2-28a

22a^2-21a^2+28a - 44a = 0

a^2 + 28a - 44a = 0

a^2 -16a = 0

a(a-16) = 0

a = 0 (which cannot be)

or a-16 = 0

a = 16

Recall;

b = 3a/2

b = 3(16)/2

b = 24

step-by-step explanation: where's the phrase?

mx+89b(-98)nb

step-by-step explanation: