$A$ cable tied to the top of an electric pole is affixed at a point on the ground $a \text{ m}$ away from the pole. If the cable makes an angle $\theta$ with the ground,then prove that the height of the pole is $a \tan \theta \text{ m}$ and the length of the wire is $a \sec \theta \text{ m}$.

(N/A) Let $\overline{AB}$ be the electric pole and $\overline{AC}$ be the cable fixed at point $C$ on the ground.
In the right-angled $\Delta ABC$,$\angle B = 90^{\circ}$,$\angle C = \theta$,and the base $BC = a \text{ m}$.
To find the height of the pole $(AB)$:
Using the trigonometric ratio $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC}$.
$\tan \theta = \frac{AB}{a}$
$AB = a \tan \theta \text{ m}$.
To find the length of the cable $(AC)$:
Using the trigonometric ratio $\sec \theta = \frac{\text{Hypotenuse}}{\text{Adjacent}} = \frac{AC}{BC}$.
$\sec \theta = \frac{AC}{a}$
$AC = a \sec \theta \text{ m}$.
Thus,the height of the pole is $a \tan \theta \text{ m}$ and the length of the cable is $a \sec \theta \text{ m}$.