(a) Given \(\sin \alpha = 0.3907\)
From the sine table, \(\alpha = 23°\)
(i) \(\tan \alpha = \tan 23° = 0.4245\)
(ii) \(\cos \alpha = \cos 23° = 0.9205\)
(b) Let T be the top of the ladder and B, the bottom of the vertical wall.

From the figure, \(\sin 50° = \frac{BT}{4.5m}\)
\(BT = 4.5 \sin 50° = 4.5 \times 0.7660\)
\(BT = 3.447m\)
Distance between the top of the ladder and bottom of the window
= \(4 - 3.447 = 0.553m\)
= \( 0.553m \times 100\)
= \(55.3 cm\)