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In the adjoining figure , AM =BM and `angleB=90^(@).` If `angleBCM=theta`, then find the values of the following :
(i) `sintheta` (ii) `tan theta` (iii) `sec theta`

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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))`

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