47 Solution: Let if $11||12$ and AB is transverse to it Then, $\angle\text{PBA}$ should be equal to $\angle\text{BAS}$ (Alternate angles) So if $11||12,$ then $\angle\text{BAS}=70^\circ$ $⇒\angle\text{BAC} = 78^\circ - 35^\circ = 43^\circ\ ...\ (\text{i})$ Now, in $\angle\text{ABC}$ $\text{x}^\circ + \angle\text{C} + \angle\text{BAC} = 180^\circ$ $⇒ \text{x}^\circ + 90^\circ + 43^\circ = 180^\circ$ $⇒ \text{x}^\circ = 180^\circ - 90^\circ - 43^\circ = 47^\circ$ $⇒ \text{x}^\circ = 47^\circ$ So if $\text{x}^\circ = 47^\circ$ then $11||12$
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