Answer
$$1.19$$
Work Step by Step
At the static equilibrium, the net force acting on the the climber zero.
Therefore,
at horizontal equilibrium: $F_N-T\sin\phi=0\implies F_N=T\sin\phi\;........(1)$
at vertical equilibrium: $f_s+T\cos\phi-W=0\;...............(2)$
The net torque about the contact point between the wall and her feet and is also zero.
$WL\sin\theta-TL\sin(180^{\circ}-\theta-\phi)=0$
or, $T=\frac{W\sin\theta}{\sin(180^{\circ}-\theta-\phi)}\;...........(3)$
Again,
$f_s=\mu_sF_N$
or, $f_s=\mu_sT\sin\phi$
Substituting in Eq. $2$, we obtain
$\mu_sT\sin\phi+T\cos\phi-W=0$
or, $T(\mu_s\sin\phi+\cos\phi)=W$
Using Eq. $3$, we obtain
$\frac{W\sin\theta}{\sin(180^{\circ}-\theta-\phi)}(\mu_s\sin\phi+\cos\phi)=W$
or, $\mu_s=\frac{1}{\sin\phi}\Big(\frac{\sin(180^{\circ}-\theta-\phi)}{\sin\theta}-\cos\phi\Big)$
Substituting the given values, we obtain
$\mu_s=\frac{1}{\sin30^{\circ}}\Big(\frac{\sin(180^{\circ}-40^{\circ}-30^{\circ})}{\sin40^{\circ}}-\cos30^{\circ}\Big)$
or, $\boxed{\mu_s=1.19}$
Therefore, the coefficient of static friction between her climbing shoes and the wall is $1.19$
