Viewing points $A$ and $B$ are at a horizontal distance of 36 metres and 28 metres, respectively, from a clock tower. The viewing angle to the clockface at point $B$ is $64^{\circ}$. a Find the height of the clockface above the viewing level, to 3 decimal places. b Find the viewing angle to the clockface at point $A$, to 2 decimal places.

Step 1 ：We consider point B and the clock tower to form a right triangle, with the horizontal distance from point B to the clock tower as the adjacent side, the height of the clockface above the viewing level as the opposite side, and the line from point B to the clockface as the hypotenuse. The tangent of the angle of elevation (the viewing angle) is the ratio of the opposite side to the adjacent side.

Step 2 ：We can use the tangent function to find the height of the clockface above the viewing level. The tangent of the viewing angle at point B is \(\tan(64^\circ)\), and the horizontal distance from point B to the clock tower is 28 metres. So, we have \(\tan(64^\circ) = \frac{\text{height}}{28}\).

Step 3 ：Solving for height, we get \(\text{height} = 28\tan(64^\circ)\).

Step 4 ：Calculating this gives us the height of the clockface above the viewing level to be approximately \(\boxed{60.352}\) metres, to 3 decimal places.

Step 5 ：Now, we consider point A and the clock tower to form another right triangle, with the horizontal distance from point A to the clock tower as the adjacent side, the height of the clockface above the viewing level as the opposite side, and the line from point A to the clockface as the hypotenuse. The tangent of the angle of elevation (the viewing angle) is the ratio of the opposite side to the adjacent side.

Step 6 ：We can use the tangent function to find the viewing angle at point A. The tangent of the viewing angle at point A is \(\tan(\theta)\), where \(\theta\) is the viewing angle, and the horizontal distance from point A to the clock tower is 36 metres. So, we have \(\tan(\theta) = \frac{60.352}{36}\).

Step 7 ：Solving for \(\theta\), we get \(\theta = \tan^{-1}\left(\frac{60.352}{36}\right)\).

Step 8 ：Calculating this gives us the viewing angle at point A to be approximately \(\boxed{59.04^\circ}\), to 2 decimal places.