Normal and tangential components:

If σ_x be the stress acting on the plane general to x axis (longitudinal axis), then the axial force P = σ_x × A. That force acting on the inclined plane might be resolved into normal and tangential components.

Normal component on the plane = P cos θ

= σ_x × A cos θ

Tangential component = - P sin θ

= - σ_x × A sin θ

Normal stress on the plane   = σ_x × A cos θ/(A/cos θ) = σ_x × cos²θ

Shear stress on the plane    =- σ x × Asin θ/(A/cos θ) =- σ_x × cos θ sin θ

Figure

If on the cross sectional area of the original solid shear stress τ_xy is applied, its components on the inclined plane may be evaluated as,

Normal stress component = τ_xy cos θ sin θ

Shear stress component      = τ_xy cos²θ

By a similar analysis (your exercise), you might verify the subsequent:

If a normal stress of σ_y is applied on the solid, then the stress components on a plane whose normal is inclined at θ to the x axis are given by

Normal stress = σ_y sin²θ

Shear stress   = σ_y sin θ cos θ

If a shear stress of τ_yx is applied on a solid, the stress components on the inclined plane are given by (with τ_yx = τ_xy).

Normal stress = τ_xy sin θ cos θ

Shear stress   = τ_xy sin2θ