Heights & Distances Questions and Answers

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: B

The shadow will be 5 meter long.

The tower's height is the shadow length times the tangent of the Sun's altitude:

\[ \begin{aligned} h &= 15 \tan 30^{\circ} = \frac{15}{\sqrt{3}} = 5\sqrt{3} \\ s &= \frac{h}{\tan 60^{\circ}} = \frac{5\sqrt{3}}{\sqrt{3}} = 5 \end{aligned} \]

Since \(\tan 60^{\circ} = 3 \tan 30^{\circ}\), the shadow at 60° is one-third of the shadow at 30°. 4, 3 and 6 metres do not give that ratio.

Answer: C

The height of the tree is \(16\sqrt{3}\) meter.

The tree, its shadow and the Sun's ray form a right triangle in which the angle of elevation is 60°:

\[ \begin{aligned} \tan 60^{\circ} &= \frac{h}{16} \\ h &= 16\sqrt{3} \approx 27.7 \text{ m} \end{aligned} \]

\(16/\sqrt{3}\) is the answer for an elevation of 30°, where the tree would be shorter than its shadow. 16 m would need an elevation of 45°, and 8 m is half the shadow, which corresponds to no standard angle.

Answer: A

The height of the tower is \(17/\sqrt{3}\) m.

The tower, the ground and the line of sight form a right triangle with the 30° angle at the observer, the 17 m distance as the adjacent side and the height h as the opposite side.

\[ \begin{aligned} \tan 30^{\circ} &= \frac{h}{17} \\ h &= \frac{17}{\sqrt{3}} \approx 9.8\ \text{m} \end{aligned} \]

\(17\sqrt{3}\) would be the height for a 60° angle of elevation, and 17 m would need a 45° angle.

Answer: A

The answer is x = 7.5 metre and y = \(15\sqrt{3}/2\) metre.

The ladder is the hypotenuse, 15 m, making 30° with the ground. The height reached is the side opposite that angle and the distance from the wall the side adjacent to it:

\[ \begin{aligned} x &= 15 \sin 30^\circ = 7.5\ \text{m} \\ y &= 15 \cos 30^\circ = \frac{15\sqrt{3}}{2}\ \text{m} \end{aligned} \]

Option B swaps the two, which would be right only for a 60° angle with the ground. The values with \(\sqrt{2}\) belong to a 45° angle.

Answer: C

The angle of elevation is 60°.

From the end of the shadow, the tower's height is the opposite side and the shadow the adjacent side:

\[ \begin{aligned} \tan\theta &= \frac{\text{height}}{\text{shadow}} \\ &= \frac{\sqrt{3}}{1} \end{aligned} \]

Since \(\tan 60^\circ = \sqrt{3}\), the angle is 60°. 30° would need the ratio \(1 : \sqrt{3}\), the reverse, and 45° would need a shadow equal to the height.

Answer: A

The chimney is 30 metres tall.

From the observer's eyes, the top of the chimney is at 45°, so the height above eye level equals the horizontal distance:

\[ \begin{aligned} h_{1} &= 28.5 \times \tan 45^{\circ} = 28.5\ \text{m} \\ H &= 28.5 + 1.5 = 30\ \text{m} \end{aligned} \]

The observer's eye height must be added, since the angle is measured from his eyes, not from the ground. The other options do not follow from the given distance and angle.

Answer: A

The tower is 21.6 m high.

Measure from the observer's eye, 1.6 m above the ground, at a horizontal distance of \(20\sqrt{3}\) m (the paper omits the unit):

\[ \begin{aligned} h' &= 20\sqrt{3} \tan 30^\circ \\ &= 20\sqrt{3} \times \frac{1}{\sqrt{3}} = 20\ \text{m} \\ h &= 20 + 1.6 = 21.6\ \text{m} \end{aligned} \]

Leaving out the eye height gives 20 m, which is not offered. 23.2 m adds the 1.6 m twice. 24.72 m and 20.3 m do not follow from \(\tan 30^\circ\).

Answer: B

The height is d/ (Cot α - Cot β).

Let the tower have height \(h\) and let \(x\) be the distance from A to its foot. Walking \(d\) towards the tower leaves \(x - d\), where the angle is the larger \(\beta\):

\[ \begin{aligned} x &= h \cot \alpha \\ x - d &= h \cot \beta \\ d &= h(\cot \alpha - \cot \beta) \\ h &= \frac{d}{\cot \alpha - \cot \beta} \end{aligned} \]

The sum \(\cot \alpha + \cot \beta\) applies when the two observation points are on opposite sides of the tower, so that the distances add. The options mixing tan and cot do not come from either case.

Answer: C

The height of the tower is 27.3 m.

The angle rises from 30° to 45°. Let the height be \(h\) and the nearer distance \(x\). At 45°, \(h = x\). From the first point, 20 m further away:

\[ \begin{aligned} \tan 30^\circ &= \frac{h}{x + 20} = \frac{h}{h + 20} \\ \sqrt{3}\,h &= h + 20 \\ h &= \frac{20}{\sqrt{3} - 1} = \frac{20}{0.732} = 27.3 \end{aligned} \]

The trap is reading the second angle as 15°: the question says the angle increases by 15°, so it becomes 45°. 17.3 m is \(30 \tan 30^\circ\), a mix-up of the figures given.