Heights & Distances Questions and Answers

Answer: C

The tower is opposite the \( 30^{\circ} \) angle and the distance is adjacent, so:

\[ \tan 30^{\circ} = \frac{100}{d} \quad\Longrightarrow\quad d = \frac{100}{\tan 30^{\circ}} = 100\sqrt{3} \]

\[ d = 100 \times 1.732 = 173.2 \approx 173 \text{ m} \]

A \( 30^{\circ} \) elevation always puts the observer \( \sqrt{3} \) times the height away. Multiplying by \( \tan 30^{\circ} \) instead of dividing gives about 58 m - the classic inversion.

Answer: A

The balloon ends up directly overhead at 450 m, so measure both earlier observations from that vertical line.

From A, at \( 45^{\circ} \):

\[ d_{A} = \frac{450}{\tan 45^{\circ}} = 450 \text{ m} \]

From B, at \( 60^{\circ} \):

\[ d_{B} = \frac{450}{\tan 60^{\circ}} = \frac{450}{\sqrt{3}} = 150\sqrt{3} = 259.8 \text{ m} \]

\[ AB = d_{A} - d_{B} = 450 - 150\sqrt{3} = 150\left(3 - \sqrt{3}\right) \]

\[ = 150 \times 1.268 = 190.2 \text{ m} \]

The balloon is assumed to hold its height while the man walks, so all three sightings share the same 450 m - that is what makes the two distances comparable.

Answer: C

Let the lower part be \( h \), so the upper is \( 4h \) and the pole is \( 5h \). If the lower part subtends \( \alpha \) at the observation point, the whole pole subtends \( 2\alpha \).

\[ \tan\alpha = \frac{h}{16}, \qquad \tan 2\alpha = \frac{5h}{16} \]

Writing \( t = \dfrac{h}{16} \) and using \( \tan 2\alpha = \dfrac{2t}{1-t^{2}} \):

\[ 5t = \frac{2t}{1-t^{2}} \quad\Longrightarrow\quad 5(1-t^{2}) = 2 \quad\Longrightarrow\quad t^{2} = \frac{3}{5} \]

\[ \frac{h^{2}}{256} = \frac{3}{5} \quad\Longrightarrow\quad h = 16\sqrt{\frac{3}{5}} \]

\[ \text{pole} = 5h = 80\sqrt{\frac{3}{5}} = 16\sqrt{15} \approx 61.97 \text{ m} \]

"Equal angles" is what makes the total exactly \( 2\alpha \), and that is the hook the double-angle identity hangs on.

Answer: A

The angle of depression from the masthead equals the angle of elevation of the masthead from the boat, so the 210 m height is opposite the \( 30^{\circ} \).

\[ \tan 30^{\circ} = \frac{210}{d} \]

\[ d = \frac{210}{\tan 30^{\circ}} = 210\sqrt{3} \approx 363.7 \text{ m} \]

At \( 30^{\circ} \) the horizontal distance is always \( \sqrt{3} \) times the height, so it must be larger than 210 m - which rules out \( 70\sqrt{3} \) and \( 105\sqrt{3} \) at a glance.

Answer: A

The ladder is the hypotenuse and the 3 m is adjacent to the \( 45^{\circ} \) angle, so cosine relates them.

\[ \cos 45^{\circ} = \frac{3}{L} \quad\Longrightarrow\quad \frac{1}{\sqrt{2}} = \frac{3}{L} \]

\[ L = 3\sqrt{2} = 3 \times 1.414 = 4.243 \approx 4.23 \text{ m} \]

At \( 45^{\circ} \) the ladder reaches 3 m up the wall as well, so it is the diagonal of a 3 m square - hence the factor \( \sqrt{2} \).

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: A

The angle of elevation \( \theta \) satisfies

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

With the shadow \( \sqrt{3} \) times the height \( h \):

\[ \tan\theta = \frac{h}{\sqrt{3}\,h} = \frac{1}{\sqrt{3}} \quad\Longrightarrow\quad \theta = 30^{\circ} \]

The pole's height cancels, so no measurement is needed. A shadow longer than the object always means the sun is below \( 45^{\circ} \).

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 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.