The problem consists of two parts: (a) From the top $K$ of a building 320 m high, the angles of depression of the top $L$ and bottom $M$ of another building on the same horizontal ground are $29^\circ$ and $41^\circ$ respectively. We are asked to (i) illustrate the information in a diagram, and (ii) calculate, correct to the nearest meter, the height of the other building. (b) The time taken to travel a distance of 120 km was reduced by 30 minutes when the speed was increased by 20 km/h. We are asked to calculate the initial speed.

1. Problem Description

The problem consists of two parts:

(a) From the top $K$ of a building 320 m high, the angles of depression of the top $L$ and bottom $M$ of another building on the same horizontal ground are $29^\circ$ and $41^\circ$ respectively. We are asked to (i) illustrate the information in a diagram, and (ii) calculate, correct to the nearest meter, the height of the other building.

(b) The time taken to travel a distance of 120 km was reduced by 30 minutes when the speed was increased by 20 km/h. We are asked to calculate the initial speed.

2. Solution Steps

(a) (i) Diagram:

```

K

*

/|

/ | 29°

/ |

/ |

L----*

| |

| |

| |

M----*

| |

| |

| | 320m

*----*

```

(a) (ii) Let the height of the taller building be $h_1 = 320$ m, and the height of the shorter building be $h_2$. Let $d$ be the horizontal distance between the two buildings.

From the angle of depression of $M$, we have:

$tan(41^\circ) = \frac{320}{d}$

$d = \frac{320}{tan(41^\circ)}$

From the angle of depression of $L$, we have:

$tan(29^\circ) = \frac{320 - h_2}{d}$

$d = \frac{320 - h_2}{tan(29^\circ)}$

Therefore:

$\frac{320}{tan(41^\circ)} = \frac{320 - h_2}{tan(29^\circ)}$

$320 \cdot tan(29^\circ) = (320 - h_2) \cdot tan(41^\circ)$

$320 \cdot tan(29^\circ) = 320 \cdot tan(41^\circ) - h_2 \cdot tan(41^\circ)$

$h_2 \cdot tan(41^\circ) = 320 \cdot tan(41^\circ) - 320 \cdot tan(29^\circ)$

$h_2 = \frac{320 \cdot (tan(41^\circ) - tan(29^\circ))}{tan(41^\circ)}$

$h_2 = 320 \cdot (1 - \frac{tan(29^\circ)}{tan(41^\circ)})$

Using a calculator:

$tan(29^\circ) \approx 0.5543$

$tan(41^\circ) \approx 0.8693$

$h_2 \approx 320 \cdot (1 - \frac{0.5543}{0.8693})$

$h_2 \approx 320 \cdot (1 - 0.6376)$

$h_2 \approx 320 \cdot 0.3624$

$h_2 \approx 115.968$

The height of the other building is approximately 116 m.

(b) Let $v$ be the initial speed in km/h.

The initial time taken is $t_1 = \frac{120}{v}$.

The new speed is $v+20$ km/h.

The new time taken is $t_2 = \frac{120}{v+20}$.

The time is reduced by 30 minutes, which is 0.5 hours.

$t_1 - t_2 = 0.5$

$\frac{120}{v} - \frac{120}{v+20} = 0.5$

$120(\frac{1}{v} - \frac{1}{v+20}) = 0.5$

$120(\frac{v+20-v}{v(v+20)}) = 0.5$

$120(\frac{20}{v(v+20)}) = 0.5$

$\frac{2400}{v(v+20)} = 0.5$

$2400 = 0.5v(v+20)$

$4800 = v(v+20)$

$v^2 + 20v - 4800 = 0$

Using the quadratic formula:

$v = \frac{-20 \pm \sqrt{20^2 - 4(1)(-4800)}}{2(1)}$

$v = \frac{-20 \pm \sqrt{400 + 19200}}{2}$

$v = \frac{-20 \pm \sqrt{19600}}{2}$

$v = \frac{-20 \pm 140}{2}$

Since $v$ must be positive:

$v = \frac{-20 + 140}{2}$

$v = \frac{120}{2}$

$v = 60$ km/h

3. Final Answer

(a) (ii) 116 m

(b) 60 km/h