In $$ {\Lambda}_b^0\to {\Lambda}_c^{+}\left(\to {\Lambda}^0{\pi}^{+}\right){\tau}^{-}{\overline{v}}_{\tau } $$
Λ
b
0
→
Λ
c
+
→
Λ
0
π
+
τ
−
v
¯
τ
decay, the solid angle of the final-state particle τ− cannot be determined precisely since the decay products of the τ− include an undetected ντ. Therefore, the angular distribution of this decay cannot be measured. In this work, we construct a measurable angular distribution by considering the subsequent decay τ−→ π−ντ. The full cascade decay is $$ {\Lambda}_b^0\to {\Lambda}_c^{+}\left(\to {\Lambda}^0{\pi}^{+}\right){\tau}^{-}\left(\to {\pi}^{-}{v}_{\tau}\right){\overline{v}}_{\tau } $$
Λ
b
0
→
Λ
c
+
→
Λ
0
π
+
τ
−
→
π
−
v
τ
v
¯
τ
. The three-momenta of the final-state particles Λ0, π+, and π− can be measured. Considering all Lorentz structures of the new physics (NP) effective operators and an unpolarized initial Λb state, the five-fold differential angular distribution can be expressed in terms of ten angular observables $$ {\mathcal{K}}_i\left({q}^2,{E}_{\pi}\right) $$
K
i
q
2
E
π
. By integrating over some of the five kinematic parameters, we define a number of observables, such as the Λc spin polarization $$ {P}_{\Lambda_c}\left({q}^2\right) $$
P
Λ
c
q
2
and the forward-backward asymmetry of π− meson AFB(q2), both of which can be represented by the angular observables $$ {\hat{\mathcal{K}}}_i\left({q}^2\right) $$
K
̂
i
q
2
. We provide numerical results for the entire set of the angular observables $$ {\hat{\mathcal{K}}}_i\left({q}^2\right) $$
K
̂
i
q
2
and $$ {\hat{\mathcal{K}}}_i $$
K
̂
i
both within the Standard Model and in some NP scenarios, which are a variety of best-fit solutions in seven different NP hypotheses. We find that the NP which can resolve the anomalies in $$ \overline{B}\to {D}^{\left(\ast \right)}{\tau}^{-}{\overline{v}}_{\tau } $$
B
¯
→
D
∗
τ
−
v
¯
τ
decays has obvious effects on the angular observables $$ {\hat{\mathcal{K}}}_i\left({q}^2\right) $$
K
̂
i
q
2
, except $$ {\hat{\mathcal{K}}}_{1 ss}\left({q}^2\right) $$
K
̂
1
ss
q
2
and $$ {\hat{\mathcal{K}}}_{1 cc}\left({q}^2\right) $$
K
̂
1
cc
q
2
.