Heights & Distances Questions and Answers

Answer: A

Let the cloud be \( H \) m above the lake surface, and let \( d \) be the horizontal distance from the observation point. The observer is 200 m above the lake, and the reflection sits \( H \) m below the surface.

Relative to the observer, the cloud is \( H - 200 \) up and the reflection is \( H + 200 \) down:

\[ \tan 30^{\circ} = \frac{H - 200}{d}, \qquad \tan 60^{\circ} = \frac{H + 200}{d} \]

Dividing eliminates \( d \):

\[ \frac{\tan 60^{\circ}}{\tan 30^{\circ}} = \frac{\sqrt{3}}{1/\sqrt{3}} = 3 = \frac{H+200}{H-200} \]

\[ H + 200 = 3H - 600 \quad\Longrightarrow\quad 2H = 800 \quad\Longrightarrow\quad H = 400 \text{ m} \]

The reflection is the mirror image in the water surface, so its depth below the surface equals the cloud's height above it - that is why the second numerator is \( H + 200 \) and not \( H - 200 \).

Answer: B

The swimmer's path is the hypotenuse. The width of the river is the perpendicular distance between the parallel banks, which is the side opposite the \( 45^{\circ} \) angle the path makes with the bank.

\[ \sin 45^{\circ} = \frac{\text{width}}{20} \]

\[ \text{width} = 20 \times \frac{1}{\sqrt{2}} = 10\sqrt{2} = 10 \times 1.414 = 14.14 \text{ m} \]

At \( 45^{\circ} \) the swimmer drifts exactly as far along the bank as the river is wide, so the 20 m swim splits into two equal legs of \( 10\sqrt{2} \) m.

Answer: C

Let the cliff be \( h \). Depression from the top equals elevation from the trawler, so:

\[ d_{30} = \frac{h}{\tan 30^{\circ}} = h\sqrt{3}, \qquad d_{60} = \frac{h}{\tan 60^{\circ}} = \frac{h}{\sqrt{3}} \]

In 6 minutes the trawler covers

\[ h\sqrt{3} - \frac{h}{\sqrt{3}} = \frac{3h - h}{\sqrt{3}} = \frac{2h}{\sqrt{3}} \]

and the distance still to go is \( \dfrac{h}{\sqrt{3}} \), exactly half of that. At constant speed the remaining time is therefore half of 6 minutes:

\[ t = 6 \times \frac{h/\sqrt{3}}{2h/\sqrt{3}} = 3 \text{ minutes} \]

The cliff height cancels, so it is never needed.

Answer: A

The cliff is 150 m, and the angle of depression falls as the boat moves away.

\[ d_{60} = \frac{150}{\tan 60^{\circ}} = \frac{150}{\sqrt{3}} = 50\sqrt{3} = 86.6 \text{ m} \]

\[ d_{45} = \frac{150}{\tan 45^{\circ}} = 150 \text{ m} \]

\[ \text{distance rowed} = 150 - 86.6 = 63.4 \text{ m in } 2 \text{ minutes} \]

\[ \text{speed} = \frac{63.4}{2} = 31.7 \text{ m/min} = 31.7 \times \frac{60}{1000} = 1.90 \text{ km/h} \]

Converting metres per minute to km/h means multiplying by 60 and dividing by 1000 - the same as multiplying by 0.06.

Answer: D

The tower is opposite the angle of elevation and the 100 m is the adjacent side, so the tangent gives the angle directly.

\[ \tan\theta = \frac{100\sqrt{3}}{100} = \sqrt{3} \]

\[ \theta = 60^{\circ} \]

The three standard tangents are worth knowing by sight: \( \tan 30^{\circ} = \tfrac{1}{\sqrt{3}} \), \( \tan 45^{\circ} = 1 \) and \( \tan 60^{\circ} = \sqrt{3} \). Since the tower is taller than the distance, the angle must exceed \( 45^{\circ} \).

Answer: B

Both sightings are taken from the same point 40 m away, so they share one base.

Tower, at \( 30^{\circ} \):

\[ h = 40\tan 30^{\circ} = \frac{40}{\sqrt{3}} = 23.09 \approx 23.1 \text{ m} \]

Top of the tank, at \( 45^{\circ} \):

\[ H = 40\tan 45^{\circ} = 40 \text{ m} \]

The tank stands on the tower, so its own height is the difference:

\[ 40 - 23.09 = 16.91 \approx 16.9 \text{ m} \]

The second angle measures to the top of the tank, giving the combined height - reporting 40 m as the tank's height is the error the distractors target.

Answer: C

P and Q are directly opposite, so the two horizontal distances from the tree add up to the river's full width of 100 m.

With the tree of height \( h \):

\[ d_{30} = \frac{h}{\tan 30^{\circ}} = h\sqrt{3}, \qquad d_{45} = \frac{h}{\tan 45^{\circ}} = h \]

\[ h\sqrt{3} + h = 100 \quad\Longrightarrow\quad h(\sqrt{3}+1) = 100 \]

\[ h = \frac{100}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1} = \frac{100(\sqrt{3}-1)}{2} = 50(\sqrt{3}-1) \]

\[ h = 50 \times 0.732 = 36.6 \text{ m} \]

The island is not at the centre: the \( 30^{\circ} \) bank is \( \sqrt{3} \) times further away than the \( 45^{\circ} \) one.

Answer: A

Place the mountain's foot at the origin. The summit is seen at \( 45^{\circ} \), so for a height \( H \) km it sits at \( (H,\;H) \).

Ascending 1000 m = 1 km up a \( 30^{\circ} \) slope moves the observer to

\[ \left(\cos 30^{\circ},\; \sin 30^{\circ}\right) = \left(\frac{\sqrt{3}}{2},\; \frac{1}{2}\right) \]

From there the elevation is \( 60^{\circ} \):

\[ \frac{H - \tfrac{1}{2}}{H - \tfrac{\sqrt{3}}{2}} = \sqrt{3} \]

\[ H - \frac{1}{2} = \sqrt{3}H - \frac{3}{2} \quad\Longrightarrow\quad H(\sqrt{3}-1) = 1 \]

\[ H = \frac{1}{\sqrt{3}-1} = \frac{\sqrt{3}+1}{2} = 1.366 \text{ km} \]

Climbing a slope raises the observer as well as bringing them nearer, so both coordinates change.

Answer: C

Both planes lie on the same vertical line, so one horizontal distance \( d \) serves both sightings. The higher plane takes the steeper angle.

\[ \tan 60^{\circ} = \frac{4000}{d} \quad\Longrightarrow\quad d = \frac{4000}{1.732} = 2309.47 \text{ m} \]

\[ h = d\tan 45^{\circ} = 2309.47 \text{ m} \]

\[ \text{vertical gap} = 4000 - 2309.47 = 1690.53 \text{ m} \]

At \( 45^{\circ} \) the lower plane's height simply equals the horizontal distance, which is why only one division is really needed.

Answer: A

Fig.
Let AB be the hill whose foot is B and D and C are two kilometer stones. 
Therefore DC = 1 km = 1000 m,
 Let AB = h and BC = x.
In right angled Δ ABC,
tan 45º = AB/BC
=> 1 = h/x
=> x = h                                                       ------------- (1) 
In right angled Δ ABC,
 tan 30º = AB/BD
=> 1/√3 = h/(x + 1000) 
=> x + 1000 = h√3                                            ----------------(2) 
using (1) and (2) we get
x + 1000 = x√3
 => x(√3 - 1) = 1000 
Or, x = [1000/(√3 - 1)] × (√3 + 1)/(√3 + 1) 
Or, x = 1000 (√3 + 1)/2
= 500(√3 + 1)
= 500 × 2.732 
= 1366 m or 1.366 km. 
Therefore first km stone is 1.366 km and second km stone is 2.366 km from the foot of the hill.