i have some question about finding the values of the voltage divider configuration of a bjt...
Why is voltage emitter is one tenth of Vcc?
Why is the current in the second resistor in the base is 10 times Ib?
kindly please give me the right explanation why are this values considered like this...THANK YOU IN ADVANCE

 

Why is voltage emitter is one tenth of Vcc?

This is the rule of thumb. For good thermal stability we choose Ve to be grater then Vbe.
So for good thermal stability we usually choose Re=(0.1 ... 0.4)*Vcc/IC or Ve large then 1V (Ve>>Vbe)

Why is the current in the second resistor in the base is 10 times Ib?

We have very similarly situation with 10xIb rule of the thumb.
Becaues current gain of a BJT is not constant, beta change with Ic and temperature.
And we want that voltage divider set the proper Ic current.
To achieve this we select the current drawn from a divider is a small fraction of the current through the resistors, then the voltage divider action is not upset very much. That's why you make the current in the divider resistors 10 times the base current. Then you can effectively ignore the upsetting effect of the base current on the divider action.