# Please solve the 19th problem. Please do not provide any link. Q.19.  The following figure shows a triangle ABC with exterior angles as x, y and z. (i) If AB > AC > BC; arrange the angles x, y and z in ascending order of their values. (ii) In the same figure, if y > x > z; arrange sides AB, BC and AC in descending order of their lengths.

Dear Student,

Here , Exterior$\angle$ BAC + $\angle$ BAC = 180$°$                         ( Linear pair angles )
So,

x  + $\angle$ BAC = 180$°$   , By substituting value from given diagram

$\angle$ BAC  = 180$°$ - x

Similarly we get : $\angle$ ABC  = 180$°$ - y   and $\angle$ ACB  = 180$°$ - z

Then ,

$\angle$ BAC  = 180$°$ - x  , $\angle$ ABC  = 180$°$ - y   and $\angle$ ACB  = 180$°$ - z                      --- ( 1 )

i ) Given :  AB > AC > BC

And we know theorem : In two unequal sides of triangle , the angle opposite to longer side is larger vice versa .

So ,

$\angle$ ACB > $\angle$ ABC >  $\angle$ BAC , Substitute values from equation 1 we get :

180$°$ - z > 180$°$ - y  > 180$°$ - x  , Now we subtract 180$°$ and get :

- z > - y > - x  , Now we multiply by ' - 1 ' and get :

z  <  y < x                                                                       ( Ans )

ii ) Given : y > xz  , Now we multiply by ' - 1 ' and get :

- y < - x  < - , Now we add 180$°$ and get :

180$°$- y < 180$°$- x  < 180$°$- , Substitute values from equation 1 we get :

$\angle$ ABC <  $\angle$ BAC < $\angle$ ACB , By using above mention theorem we get :

AC < BC < AB

Then we write it in descending order , As :

AB > BC > AC                                                        ( Ans )