Heights & Distances Questions and Answers

Answer: D

Join the two tops. That line's vertical leg is the height difference and its horizontal leg is the gap between the poles.

\[ \text{height difference} = 80 - 62.5 = 17.5 \text{ m} \]

\[ \tan 45^{\circ} = \frac{17.5}{d} = 1 \quad\Longrightarrow\quad d = 17.5 \text{ m} \]

At \( 45^{\circ} \) the two legs are always equal, so the distance simply is the height difference. Using either height on its own, or their sum, is where the other options come from.

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

Let the tower be \( h \) m and the observation point \( d \) m away. The bottom of the flagstaff is the top of the tower.

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

\[ \tan 45^{\circ} = \frac{h+7}{d} \quad\Longrightarrow\quad h + 7 = d = h\sqrt{3} \]

\[ h(\sqrt{3} - 1) = 7 \quad\Longrightarrow\quad h = \frac{7}{\sqrt{3}-1} = \frac{7(\sqrt{3}+1)}{2} \]

\[ h = 3.5 \times 2.732 = 9.562 \text{ m} \]

Both sightings share the same \( d \), which is what allows one to be substituted into the other.

Answer: A

Height and shadow are the two legs of the right triangle formed by the sun's ray, and the angle of elevation \( \theta \) sits at the tip of the shadow.

\[ \tan\theta = \frac{\text{height of tree}}{\text{length of shadow}} \]

The two are equal, so the ratio is 1:

\[ \tan\theta = 1 \quad\Longrightarrow\quad \theta = 45^{\circ} \]

No length is needed - only the ratio matters, which is why the question can omit the tree's actual height. At \( 30^{\circ} \) the shadow would be \( \sqrt{3} \) times the height; at \( 60^{\circ} \), \( \tfrac{1}{\sqrt{3}} \) of it.

Answer: B

The string is the hypotenuse and the kite's height is opposite the \( 60^{\circ} \) inclination, so sine applies.

\[ \sin 60^{\circ} = \frac{75}{L} \quad\Longrightarrow\quad \frac{\sqrt{3}}{2} = \frac{75}{L} \]

\[ L = \frac{150}{\sqrt{3}} = \frac{150\sqrt{3}}{3} = 50\sqrt{3} \approx 86.6 \text{ m} \]

Rationalising \( \tfrac{150}{\sqrt{3}} \) into \( 50\sqrt{3} \) is what matches it to the listed form. The string must be longer than the 75 m height, which rules out the two divided options immediately.

Answer: B

The elevation is given as an inverse sine, so convert it to a tangent before using it with the horizontal distance.

\[ \sin\theta = \frac{3}{5} \quad\Longrightarrow\quad \cos\theta = \sqrt{1 - \left(\tfrac{3}{5}\right)^{2}} = \frac{4}{5} \]

\[ \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{3}{4} \]

\[ h = 20\tan\theta = 20 \times \frac{3}{4} = 15 \text{ m} \]

This is the 3–4–5 triangle. Using \( \tfrac{3}{5} \) directly with the 20 m base gives 12 m, the error the question is testing for.

Answer: B

Let the tower be \( T \) m high and \( d \) m away. Two sightings:

\[ \tan b = \frac{T}{d} = \frac{4}{5}, \qquad \tan a = \frac{T - 230}{d} = \frac{5}{12} \]

Subtracting eliminates \( T \):

\[ \frac{230}{d} = \frac{4}{5} - \frac{5}{12} = \frac{48 - 25}{60} = \frac{23}{60} \]

\[ d = 230 \times \frac{60}{23} = 600 \text{ m} \]

\[ T = \frac{4}{5} \times 600 = 480 \text{ m}, \qquad T - 230 = 250 \text{ m} \]

The line joining the two tops is the hypotenuse of a right triangle with legs \( d \) and \( T - 230 \):

\[ \sqrt{600^{2} + 250^{2}} = \sqrt{360000 + 62500} = \sqrt{422500} = 650 \text{ m} \]

600 m is the horizontal distance only; the question asks for the slant distance between the tops.

Answer: B

The balloon rises vertically above \( A \), so its height is the opposite side of a right triangle with base 150 m.

\[ h = 150\tan 60^{\circ} = 150\sqrt{3} = 150 \times 1.732 = 259.8 \text{ m} \]

Two minutes is 120 seconds:

\[ v = \frac{259.8}{120} = 2.165 \approx 2.16 \text{ m/s} \]

Converting the 2 minutes to seconds before dividing is the step most often skipped; leaving it in minutes gives 129.9, which matches no option.

Answer: D

Let \( d \) be the horizontal distance between the building and the tower.

The tower's foot is 30 m below the observer, at a depression of \( 45^{\circ} \):

\[ d = \frac{30}{\tan 45^{\circ}} = 30 \text{ m} \]

The tower's top, at a depression of \( 30^{\circ} \), is

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

below the building's top, so the tower's height is

\[ 30 - \frac{30}{\sqrt{3}} = 30\left(1 - \frac{1}{\sqrt{3}}\right) \approx 12.7 \text{ m} \]

The tower is shorter than the building, which is consistent with both angles being depressions rather than elevations.

Answer: D

The pole and its shadow are the two legs of the triangle formed by the sun's ray.

\[ \tan\theta = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \frac{3\sqrt{3}}{3} = \sqrt{3} \]

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

Rationalising \( \tfrac{3}{\sqrt{3}} \) to \( \sqrt{3} \) is the step that makes the standard angle visible; left as a decimal it reads 1.732, which is the same thing.