Let a = 693, b = 567 and c = 441
Now first we will find HCF of 693 and 567 by using Euclid’s division algorithm as under
693 = 567 x 1 + 126
567 = 126 x 4 + 63
126 = 63 x 2 + 0
Hence, HCF of 693 and 567 is 63
Now we will find HCF of third number i.e., 441 with 63 So by Euclid’s division alogorithm for 441 and 63
441 = 63 x 7+0
=> HCF of 441 and 63 is 63.
Hence, HCF of 441, 567 and 693 is 63.
don't quote me on this but I'm pretty sure it's H multiplication property of inequality
3. ∠1+∠2=180° and ∠2+∠3=180°
4. they are supplementary
5. ∠1+∠2=180°=∠2+∠3
6. ∠1+∠2=180°=∠2+∠3
∠2 are same in both side so ∠1=180°-∠2 =∠3=180°-∠2
7.