Inclined string on an incline plane

Abhishek, is this a problem of book Resnic Halliday (Principles of Physics).?

If we resolve mg parallel to the plane.It will come out to be

$mgsin\theta$ (in downward direction)

and if we resolve Tension in the string(T) parallel to the plane. It will come out to be

$Tcos\phi$ (in upward direction)

As sphere is at rest so

$mgsin\theta =Tcos\phi$

So

$T=\frac { 98 }{ cos(25)\sqrt { 2 } }$

On putting value of cos(25) we will get T=76.46N

It will be nice to explain with diagram.Could anyone tell me how to add diagrams in solutions?

By a free boddy diagram

$\sum_x=mg\sin45°-Tcos25°=0$

$mg \sin45°=T \cos25°$

$T=\frac{mg \sin45°}{\cos25}$

$T=\frac{98\times \sin 45°}{\cos25°} \approx \boxed{76.46}$

I did by resolving forces along vertical and horizontal. But resolving along the inclined is definitely the best solution.