A force $F$ must be acting on this plane making an angle $\left(90^{\circ}-\theta\right)$ with the normal $ON.$ Resolving $F$ into two compoents, along the plane and normal to the plane.

Component of force $\mathrm{F}$ along the plane.

$\therefore \quad \mathrm{F}_{\mathrm{P}}=\mathrm{F} \cos \theta$

component of force $\mathrm{F}$ normal to the plane.

$\mathrm{F}_{\mathrm{N}}=\mathrm{F} \cos \left(90^{\circ}-\theta\right)=\mathrm{F} \sin \theta$

Let the area of the face $\mathrm{BB}^{\prime}$ be $\mathrm{A}^{\prime}$. Then

$\frac{\mathrm{A}}{\mathrm{A}^{\prime}}=\sin \theta \quad \therefore \quad \mathrm{A}^{\prime}=\frac{\mathrm{A}}{\sin \theta}$

Tensile stress $=\frac{\mathrm{F} \sin \theta}{\mathrm{A}^{\prime}}=\frac{\mathrm{F}}{\mathrm{A}} \sin ^{2} \theta$

Shearing stress $=\frac{\mathrm{F} \cos \theta}{\mathrm{A}^{\prime}}$

$=\frac{\mathrm{F}}{\mathrm{A}} \cos \theta \sin \theta$

Their corresponding ratio is

$\frac{\text { Tensilestress }}{\text { Shearing stress }}=\frac{F}{A} \sin ^{2} \theta \times \frac{A}{F \sin \theta \cos \theta}=\tan \theta$