Heights & Distances Questions and Answers

Answer: C

Let the tower be \( h \) m and the observer \( d \) m away. The bottom of the flagpole is the top of the tower.

\[ \tan 30^{\circ} = \frac{h}{d} \quad\Longrightarrow\quad d = h\sqrt{3} \]

\[ \tan 60^{\circ} = \frac{h + 18}{d} \quad\Longrightarrow\quad h + 18 = \sqrt{3}\,d = \sqrt{3}\cdot h\sqrt{3} = 3h \]

\[ 2h = 18 \quad\Longrightarrow\quad h = 9 \text{ m} \]

Because \( \tan 60^{\circ} / \tan 30^{\circ} = 3 \), the full height is exactly three times the tower - so the flagpole is twice the tower, and 18 m makes the tower 9 m.

Answer: A

Let the lower part be \( h \), so the upper part is \( 9h \) and the pole is \( 10h \). From a point 15 m away, let \( \alpha \) be the angle subtended by the lower part. The upper part subtends the same angle, so the whole pole subtends \( 2\alpha \).

\[ \tan\alpha = \frac{h}{15}, \qquad \tan 2\alpha = \frac{10h}{15} \]

Applying the double-angle identity \( \tan 2\alpha = \dfrac{2\tan\alpha}{1-\tan^{2}\alpha} \) with \( t = \dfrac{h}{15} \):

\[ 10t = \frac{2t}{1 - t^{2}} \quad\Longrightarrow\quad 10(1 - t^{2}) = 2 \]

\[ t^{2} = \frac{8}{10} = \frac{4}{5} \quad\Longrightarrow\quad \frac{h^{2}}{225} = \frac{4}{5} \quad\Longrightarrow\quad h^{2} = 180 \]

\[ h = 6\sqrt{5} \quad\Longrightarrow\quad \text{pole} = 10h = 60\sqrt{5} \text{ m} \]

The key step is recognising that "equal angles" makes the total angle exactly double the lower one, which is what lets the double-angle formula in.

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: B

Let \( d \) be the horizontal distance from the hill to the pole.

The bottom of the pole is 100 m below the observer at a depression of 60°:

\[ \tan 60^{\circ} = \frac{100}{d} \quad\Longrightarrow\quad d = \frac{100}{\sqrt{3}} = 57.74 \text{ m} \]

The top of the pole is at a depression of 30°, so it sits

\[ d\tan 30^{\circ} = \frac{100}{\sqrt{3}} \times \frac{1}{\sqrt{3}} = \frac{100}{3} = 33.33 \text{ m} \]

below the hilltop. The pole's height is what is left:

\[ 100 - 33.33 = 66.67 \text{ m} \]

The smaller depression angle belongs to the higher point, since it is closer to the observer's eye level.

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: D

The ladder is the hypotenuse and the ground gap is the side adjacent to the 60° angle, so cosine is the ratio to use.

\[ \cos 60^{\circ} = \frac{\text{distance from wall}}{\text{ladder length}} \]

\[ \frac{1}{2} = \frac{4.6}{L} \quad\Longrightarrow\quad L = 2 \times 4.6 = 9.2 \text{ m} \]

Since \( \cos 60^{\circ} = \tfrac{1}{2} \), the ladder is always exactly twice the foot distance here. Reaching for \( \tan \) or \( \sin \) instead gives \( 4.6\sqrt{3} \) - the wall height, not the ladder.

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

The broken piece stays attached, so it becomes the hypotenuse of a right triangle whose vertical leg is the standing stump.

Let the break be at height \( h \). The fallen piece then has length \( 24 - h \), and

\[ \sin 30^{\circ} = \frac{h}{24 - h} = \frac{1}{2} \]

\[ 2h = 24 - h \quad\Longrightarrow\quad 3h = 24 \quad\Longrightarrow\quad h = 8 \text{ cm} \]

The stump and the fallen piece must add to the original 24 cm - treating the fallen piece as the full 24 cm is the common error and gives 12 cm.

Answer: C

Let the tower be \( h \) and let the car's distance be measured from its foot. Because the angle of depression from the top equals the angle of elevation from the car:

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

So in 8 minutes the car covers \( h\sqrt{3} - h = h(\sqrt{3}-1) \), and the distance still to go is \( h \). At constant speed, time is proportional to distance:

\[ t = 8 \times \frac{h}{h(\sqrt{3}-1)} = \frac{8}{\sqrt{3}-1} \]

\[ t = \frac{8(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{8(\sqrt{3}+1)}{2} = 4(\sqrt{3}+1) \]

\[ t = 4 \times 2.732 = 10.93 \text{ min} = 10 \text{ min } 56 \text{ s} \approx 10 \text{ min } 57 \text{ s} \]

The tower height cancels, so it never needed to be known.

Answer: A

Let the tower be \( h \) m and the observer \( d \) m away. The flagpole's bottom is the tower's top.

\[ \tan 30^{\circ} = \frac{h}{d} \quad\Longrightarrow\quad d = h\sqrt{3} \]

\[ \tan 60^{\circ} = \frac{h+18}{d} \quad\Longrightarrow\quad h + 18 = \sqrt{3} \cdot h\sqrt{3} = 3h \]

\[ 2h = 18 \quad\Longrightarrow\quad h = 9 \text{ m} \]

Since \( \tan 60^{\circ} \div \tan 30^{\circ} = 3 \), the tower-plus-flagpole is always three times the tower alone - so the flagpole equals twice the tower.