`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))`