Heights & Distances Questions and Answers

Answer: B

The person is now 18 ft from the tree.

The fruit hangs a quarter of the tree's height below the top: 20 - 5 = 15 ft above the ground. The stone leaves his hand at 6 ft.

\[ \begin{aligned} \text{first throw: rise} &= 20 - 6 = 14 \text{ over } 28 \\ \text{slope} &= 14/28 = 1/2 \\ \text{second throw: rise} &= 15 - 6 = 9 \\ \text{distance} &= 9 \div \tfrac{1}{2} = 18 \text{ ft} \end{aligned} \]

The same angle means the same slope, so the distance scales with the rise. 21 ft scales the 28 ft by three-quarters without allowing for the 6 ft hand height, 15 ft is the fruit's height, and 9 ft is the rise.

Answer: C

Guddi's eye, the top of the tree and the top of the mall lie on one straight line, so the tree and the mall are the vertical sides of two similar right triangles sharing the apex at Guddi.

\[ \frac{\text{tree height}}{\text{distance to tree}} = \frac{\text{mall height}}{\text{distance to mall}} \]

\[ \frac{10}{20} = \frac{H}{1000} \]

\[ H = 1000 \times \frac{10}{20} = 1000 \times 0.5 = 500 \text{ m} \]

The 1000 m is measured from Guddi, not from the tree, so it is the full base of the larger triangle - using 980 m here is the usual slip.

Answer: A

The leaning tree is the hypotenuse of a right triangle in each case, with the tree's base fixed and its 15 m length unchanged.

Towards A: the top rests 12 m up the wall, so the horizontal gap is

\[ \sqrt{15^{2} - 12^{2}} = \sqrt{225 - 144} = \sqrt{81} = 9 \text{ m} \]

Towards B: the top rests 9 m up, so

\[ \sqrt{15^{2} - 9^{2}} = \sqrt{225 - 81} = \sqrt{144} = 12 \text{ m} \]

The tree stands between the two houses, so the distance between them is the sum:

\[ 9 + 12 = 21 \text{ m} \]

Both are 9-12-15 triangles, the 3-4-5 triple scaled by three.

Answer: D

Join the two chimney tops. The horizontal leg of that line is the gap between the houses, and its vertical leg is the difference in their heights:

\[ 25 - 10 = 15 \text{ m} \]

The line makes 45° with the ground, and \( \tan 45^{\circ} = 1 \), so the two legs are equal:

\[ \tan 45^{\circ} = \frac{15}{d} = 1 \quad\Longrightarrow\quad d = 15 \text{ m} \]

The trap is using the sum 35 m or one height alone; only the height difference lies opposite the 45° angle.

Answer: C

Correct answer: 13 m

The line joining the tops is the hypotenuse of a right triangle whose horizontal side is the gap and whose vertical side is the difference in heights.

\[d = \sqrt{12^2 + (11-6)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \text{ m}\]

It is the difference of the heights that matters, not their sum.

Answer: C

The sun's rays hit the man and the pole at the same angle, so the two object–shadow triangles are similar and the ratio of height to shadow is the same for both.

\[ \frac{\text{man's height}}{\text{man's shadow}} = \frac{\text{pole's height}}{\text{pole's shadow}} \]

\[ \frac{6}{4} = \frac{h}{36} \]

\[ h = 36 \times \frac{6}{4} = 36 \times 1.5 = 54 \text{ ft} \]

The same result by scale factor: the pole's shadow is \( 36 \div 4 = 9 \) times the man's, so the pole is 9 times his height, \( 9 \times 6 = 54 \) ft. Multiplying 6 by 6 instead treats the shadow ratio as \( 36 \div 6 \), which mixes a height with a length.

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: D

The tower, its shadow and the sun's ray form a right triangle, with the angle of elevation \( \theta \) at the tip of the shadow.

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

Here the height equals the shadow, so the ratio is 1:

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

It is worth memorising the three standard cases: shadow longer than the height means the sun is low (\( 30^{\circ} \) when the shadow is \( \sqrt{3} \) times the height), equal means \( 45^{\circ} \), and shadow shorter means the sun is high (\( 60^{\circ} \) when the height is \( \sqrt{3} \) times the shadow).

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: 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.