Question

plz ans it fast it's urgent

Answer 1

wiquestionwhich question

Answer 2

Question 7,

We know, for any polygon,
sum of exterior angles is 360°

therefore,
(6x-1)+(10x+2)+(8x+2)+(9x-3)+(5x+4)+(12x+6) = 360
6x-1+10x+2+8x+2+9x-3+5x+4+12x+6= 360
50x + 10 = 360
50x = 360-10
50x = 350

$x = \frac{350}{50}$
x = 7

1st angle = 6×7-1 = 41°
2nd angle= 10×7+2 = 72°
3rd angle = 8×7+2 = 58°
4th angle = 9×7-3 = 60°
5th angle = 5×7+4 = 39°
6th angle = 12×7+6= 90°



Question 8

sum of interior angles in a pentagon= (n-2)×180°
= 5-2×180°=540°

let the common factor be x
1st angle = 4x
2nd angle= 5x
3rd angle = 6x
4th angle = 7x
5th angle = 5x

4x + 5x + 6x + 7x + 5x = 540
27x = 540

$x = \frac{540}{27}$
x = 20

1st angle = 4x = 4 × 20° = 80°
2nd angle= 5x = 5 × 20° = 100°
3rd angle = 6x = 6 × 20° = 120°
4th angle = 7x = 7 × 20° = 140°
5th angle = 5x = 5 × 20° = 100°