Part (a): Find the height of a building, given the angle of depression from the top of the building to a point on the ground and the distance from that point to the foot of the building. Part (b): Given that $PT || SU$, $QS || TR$, $|SR| = 6$ cm, $|RU| = 10$ cm, and the area of triangle $TRU = 45$ cm$^2$, calculate the area of trapezium $QTUS$.

1. Problem Description

Part (a): Find the height of a building, given the angle of depression from the top of the building to a point on the ground and the distance from that point to the foot of the building.

Part (b): Given that $PT || SU$, $QS || TR$, $|SR| = 6$ cm, $|RU| = 10$ cm, and the area of triangle $TRU = 45$ cm$^2$, calculate the area of trapezium $QTUS$.

2. Solution Steps

Part (a):

Let $h$ be the height of the building. The angle of depression from the top $T$ of the building to the point $P$ on the ground is $23.6^\circ$. This means the angle of elevation from $P$ to $T$ is also $23.6^\circ$. The distance from $P$ to the foot of the building is 50 m. We can use the tangent function to relate the angle of elevation, the height of the building, and the distance from $P$ to the foot of the building:

$$\tan(23.6^\circ) = \frac{h}{50}$$

$$h = 50 \tan(23.6^\circ)$$

$$h \approx 50(0.43697) \approx 21.8485$$

To the nearest meter, the height of the building is 22 meters.

Part (b):

The area of triangle $TRU$ is given by $\frac{1}{2} \times \text{base} \times \text{height}$. Let $h$ be the height of the triangle $TRU$ with base $RU$.

$$\text{Area of } \triangle TRU = \frac{1}{2} \times |RU| \times h = 45$$

Since $|RU| = 10$ cm,

$$\frac{1}{2} \times 10 \times h = 45$$

$$5h = 45$$

$$h = 9 \text{ cm}$$

Since $QS || TR$, the height of the trapezium $QTUS$ is equal to the height of $\triangle TRU$, which is 9 cm. The parallel sides of the trapezium $QTUS$ are $QT$ and $SU$.

Also since $PT || SU$ and $QS || TR$, $QTSR$ is a parallelogram, which means $QT = SR = 6$ cm.

Also, $SU = SR + RU = 6 + 10 = 16$ cm.

The area of trapezium $QTUS$ is given by

$$\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$$

$$\text{Area} = \frac{1}{2} \times (QT + SU) \times h$$

$$\text{Area} = \frac{1}{2} \times (6 + 16) \times 9$$

$$\text{Area} = \frac{1}{2} \times 22 \times 9$$

$$\text{Area} = 11 \times 9 = 99 \text{ cm}^2$$

3. Final Answer

Part (a): The height of the building is 22 m.

Part (b): The area of trapezium $QTUS$ is 99 cm$^2$.