From Pythagoras theorem

` BC^(2)=AC^(2)-AB^(2)=b^(2)-(2a)^(2)=b^(2)-4a^(2)`

`rArr BC=sqrt(b^(2)-4a^(2))`

Now , BM` = (AB)/(2)=`

In `DeltaBCM` ,

From Pythagoras theorem

` CM^(2)=BC^(2)+BM^(2)=(b^(2)-4a^(2))+a^(2)=b^(2)-3a^(2)`

`rArr "CM=sqrt(b^(2)-3a^(2))`

(i) ` sintheta=(BM)/(CM)=(a)/(sqrt(b^(2)-3a^(2))`

(ii)` " " tantheta=(BM)/(CM)=(a)/(sqrt(b^(2)-3a^(2))`

(iii)`sectheta=(CM)/(BC)=(sqrt(b^(2)-3a^(2)))/(sqrt(b^(2)-4a^(2))`