Correct Answer - Option 4 : 43°, 66.5°

__Concept:__

In a triaxial test,

\({\bar σ _1} = \;{\bar σ _3}{\tan ^2}\left( {45 + \frac{\phi }{2}} \right) + 2c\tan \left( {45 + \frac{\phi }{2}} \right)\)

Where σ_{1} = Major Principal stress, σ _{3} = Minor principal stress

σ̅_{3} = confining pressure

σ̅_{1} = σ̅_{3} + σ̅_{d}

σ̅_{d} = deviator stress

c = Effective cohesion

ϕ = angle of internal friction.

α = inclination of failure plane to major principal plane.

\(\alpha = 45^\circ + \frac{\phi }{2}\)

**Calculation:**

Given,

Dry sand is given, so c = 0. u = 10 kPa

σ_{d} = 218 kN/m^{2}, σ_{c} = 61 kN/m^{2}

σ_{ 1} = 61 + 218 = 279 kN/m^{2}

σ _{3} = 61 kN/m2

σ̅_{3} = 61 - 10 = 51 kN/m^{2}, σ̅_{1} = 279 - 10 = 269 kN/m^{2}.

\({\bar σ _1} = {\bar σ _3}{\tan ^2}\left( {45 + \frac{\phi }{2}} \right)\)

\(269 = 51 \times {\tan ^2}\left( {45 + \frac{\phi }{2}} \right)\)

\(45^\circ + \frac{\phi }{2} = 66.41^\circ \)

\(\frac{\phi }{2} = 21.41^\circ \)

ϕ = 42.82° ≈ 43°

\(\alpha = 45^\circ + \frac{{43}}{2} = 66.5^\circ \)