Heights & Distances Questions and Answers

Answer: A

The tower, the ground and the line of sight form a right triangle with the 30° angle at \( P \).

\[ \tan 30^{\circ} = \frac{\text{height}}{\text{distance}} \quad\Longrightarrow\quad \text{distance} = \frac{\text{height}}{\tan 30^{\circ}} \]

\[ d = \frac{200}{1/\sqrt{3}} = 200\sqrt{3} \]

\[ d = 200 \times 1.732 = 346.4 \text{ m} \]

A 30° elevation always puts the observer \( \sqrt{3} \) times the height away. Dividing by \( \sqrt{3} \) instead of multiplying gives about 115 m, which is the classic sign error.

Answer: B

The bridge stands between the banks, so the river's width is the sum of the two horizontal distances from the point on the bridge.

\[ d_{30} = \frac{2.5}{\tan 30^{\circ}} = 2.5\sqrt{3} = 2.5 \times 1.732 = 4.33 \text{ m} \]

\[ d_{45} = \frac{2.5}{\tan 45^{\circ}} = 2.5 \text{ m} \]

\[ \text{width} = 4.33 + 2.5 = 6.83 \text{ m} \]

Because the banks are on opposite sides the distances add; a bridge overhanging one bank would call for the difference instead.

Answer: D

The tree is taller than the wall, which is why the wall's top is seen below the horizontal from the tree top.

From the tree's bottom, the wall top is at 30°:

\[ \tan 30^{\circ} = \frac{11}{d} \quad\Longrightarrow\quad d = 11\sqrt{3} = 19.05 \text{ m} \]

From the tree's top, the wall top is \( H - 11 \) below, at 60°:

\[ \tan 60^{\circ} = \frac{H - 11}{11\sqrt{3}} \quad\Longrightarrow\quad H - 11 = \sqrt{3} \times 11\sqrt{3} = 33 \]

\[ H = 33 + 11 = 44 \text{ m} \]

The tree turns out to be exactly four times the wall's height - a consequence of \( \tan 60^{\circ}\tan 30^{\circ} = 1 \) giving the neat factor of 3.

Answer: A

Angles are measured from the man's eye, which is 180 cm = 1.8 m above the ground, and he is 5 m from the wall.

Bottom of the window, at 45°:

\[ \text{rise} = 5\tan 45^{\circ} = 5 \text{ m} \quad\Longrightarrow\quad \text{height} = 1.8 + 5 = 6.8 \text{ m} \]

Top of the window, at 60°:

\[ \text{rise} = 5\tan 60^{\circ} = 5\sqrt{3} = 8.66 \text{ m} \quad\Longrightarrow\quad \text{height} = 1.8 + 8.66 = 10.46 \text{ m} \]

\[ \text{window length} = 10.46 - 6.8 = 3.66 \approx 3.65 \text{ m} \]

The 1.8 m appears in both heights and cancels in the subtraction, so the eye height does not actually affect the window's length here - but it is needed if the question asks for either height on its own.

Answer: B

The men are on opposite sides, so the tower's foot lies between them and their distances add.

\[ d_{30} = \frac{50}{\tan 30^{\circ}} = 50\sqrt{3} = 50 \times 1.732 = 86.6 \text{ m} \]

\[ d_{45} = \frac{50}{\tan 45^{\circ}} = 50 \text{ m} \]

\[ \text{distance apart} = 86.6 + 50 = 136.6 \approx 136.5 \text{ m} \]

The man at \( 30^{\circ} \) stands further off; the steeper the angle, the closer the observer.

Answer: C

The lighthouse foot lies between the ships, so the total distance is the sum of the two horizontal distances.

\[ d = \frac{h}{\tan 30^{\circ}} + \frac{h}{\tan 45^{\circ}} \]

With \( h = 100 \), \( \tan 30^{\circ} = \tfrac{1}{\sqrt{3}} \) and \( \tan 45^{\circ} = 1 \):

\[ d = 100\sqrt{3} + 100 = 173.2 + 100 = 273.2 \approx 273 \text{ m} \]

Because the ships are on opposite sides the distances add. Had they been on the same side the answer would be the difference, 73 m. Choosing 173 m means only the \( 30^{\circ} \) ship was counted.

Answer: C

The ladder is the hypotenuse; the 12.4 m ground gap is adjacent to the \( 60^{\circ} \) elevation, so cosine relates them.

\[ \cos 60^{\circ} = \frac{12.4}{L} \quad\Longrightarrow\quad \frac{1}{2} = \frac{12.4}{L} \]

\[ L = 2 \times 12.4 = 24.8 \text{ m} \]

Because \( \cos 60^{\circ} = \tfrac{1}{2} \), the ladder is always exactly twice the foot distance at this angle. Using \( \tan \) instead returns \( 12.4\sqrt{3} = 21.5 \) m, which is the height reached up the wall.

Answer: C

The tower, the ground and the line of sight form a right triangle with the 30° angle at \( P \).

\[ \tan 30^{\circ} = \frac{\text{height}}{\text{distance}} \quad\Longrightarrow\quad \text{distance} = \frac{\text{height}}{\tan 30^{\circ}} \]

\[ d = \frac{100}{1/\sqrt{3}} = 100\sqrt{3} \]

\[ d = 100 \times 1.732 = 173.2 \text{ m} \]

A 30° elevation always puts the observer \( \sqrt{3} \) times the height away. Dividing by \( \sqrt{3} \) instead of multiplying gives about 58 m, which is the classic sign error.

Answer: A

The \( 60^{\circ} \) is measured from the wall, so the ground distance is opposite that angle and sine is the right ratio.

\[ \sin 60^{\circ} = \frac{x}{10} \quad\Longrightarrow\quad x = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \]

\[ x = 5 \times 1.73 = 8.65 \text{ m} \]

Reading the angle as being with the ground gives \( 10\cos 60^{\circ} = 5 \) m instead - a different problem entirely.

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.