How many bond angles of $90°$ are present in trigonal bipyramidal shape of ${\mathrm{PCl}}_5$?

${\mathrm{Cl}}_{\left(\mathrm{a}\right) }\hat{\mathrm{P}} {\mathrm{Cl}}_{\left(\mathrm{e}\right)}= 90°$
Total $\left(6\right)$ B.A.'s of$90°$ are present in T.B.P. shape of ${\mathrm{PCl}}_5$
${\mathrm{Cl}}_{\left(\mathrm{a}1\right)} \hat{\mathrm{P}} {\mathrm{Cl}}_{\left(\mathrm{e}1\right)}$             ${\mathrm{Cl}}_{\left(\mathrm{a}2\right)} \hat{\mathrm{P}} {\mathrm{Cl}}_{\left(\mathrm{e}1\right)}$
${\mathrm{Cl}}_{\left(\mathrm{a}1\right)} \hat{\mathrm{P}} {\mathrm{Cl}}_{\left(\mathrm{e}2\right)}$             ${\mathrm{Cl}}_{\left(\mathrm{a}2\right)} \hat{\mathrm{P}} {\mathrm{Cl}}_{\left(\mathrm{e}2\right)}$ 

${\mathrm{Cl}}_{\left(\mathrm{a}1\right)} \hat{\mathrm{P}} {\mathrm{Cl}}_{\left(\mathrm{e}3\right)}$             ${\mathrm{Cl}}_{\left(\mathrm{a}2\right)} \hat{\mathrm{P}} {\mathrm{Cl}}_{\left(\mathrm{e}3\right)}$