`sinA=(3)/(4)` Construct `DeltaABC` , right - angled at B such that
`BC=3kandAC=4k`
In `DeltaABC`,
From Pythagoras theorem
`AB^(2)=AC^(2)-BC^(2)=(4k)^(2)-(3k)^(2)=16k^(2)=9k^(2)=7k^(2)`
`rArrAB=sqrt(7)k`
Now , `cosA=(AB)/(AC)=(sqrt(2)k)/(4k)=(sqrt7)/(4)`
`tan A=(BC)/(AB)=(3k)/(sqrt(7)k)=(3)/(sqrt(7))`
