Heights & Distances Questions and Answers

Answer: D

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

From the tree's bottom, the wall top is at \( 30^{\circ} \):

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

From the tree's top it is \( H - 11 \) below, at \( 60^{\circ} \):

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

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

Because \( \tan 60^{\circ}\tan 30^{\circ} = 1 \), the tree works out at exactly four times the wall's height.

Answer: C

Let the poles be \( h \) and \( 2h \). The observer is at the midpoint, so each is 100 m away.

Complementary elevations \( \theta \) and \( 90^{\circ}-\theta \) have tangents whose product is 1, since \( \tan(90^{\circ}-\theta) = \cot\theta \):

\[ \frac{h}{100} \times \frac{2h}{100} = 1 \quad\Longrightarrow\quad 2h^{2} = 10000 \]

\[ h^{2} = 5000 \quad\Longrightarrow\quad h = 50\sqrt{2} = 70.71 \text{ m} \]

\[ 2h = 141.42 \text{ m} \]

So roughly 70.5 m and 141 m. Neither angle is ever needed - the complementarity alone pins the answer down.

Answer: C

The angles are taken from the man's eye, 180 cm = 1.8 m above the ground, and he stands 5 m from the wall.

Window bottom at \( 45^{\circ} \):

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

Window top at \( 60^{\circ} \):

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

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

The eye height appears in both and cancels in the subtraction, so it affects the individual heights but not the window's length.

Answer: A

The taut thread is the hypotenuse and the kite's height is the side opposite the \( 30^{\circ} \) elevation, so sine applies.

\[ \sin 30^{\circ} = \frac{h}{120} \]

\[ h = 120 \times \frac{1}{2} = 60 \text{ m} \]

Using \( \tan 30^{\circ} \) would treat 120 m as the ground distance and give \( 40\sqrt{3} \approx 69.3 \) m - but the thread is the slant length, not the base.

Answer: B

The bent tree splits into a standing stump \( h \) and a bent-over part \( L \), which together made up the original height. The 8 m is the distance from the foot to where the top touches the ground.

\[ \cos 60^{\circ} = \frac{8}{L} \quad\Longrightarrow\quad L = \frac{8}{1/2} = 16 \text{ m} \]

\[ \tan 60^{\circ} = \frac{h}{8} \quad\Longrightarrow\quad h = 8\sqrt{3} \text{ m} \]

\[ \text{original height} = h + L = 8\sqrt{3} + 16 = 8(\sqrt{3} + 2) \text{ m} \]

Numerically that is about 29.9 m. The stump and the bent part must be added - reporting \( 8\sqrt{3} \) alone gives only the part still upright.

Answer: D

Let the tower be \( h \). The two positions give:

\[ d_{30} = \frac{h}{\tan 30^{\circ}} = h\sqrt{3}, \qquad d_{60} = \frac{h}{\tan 60^{\circ}} = \frac{h}{\sqrt{3}} \]

The 100 m walked is the difference:

\[ h\sqrt{3} - \frac{h}{\sqrt{3}} = 100 \quad\Longrightarrow\quad \frac{3h - h}{\sqrt{3}} = 100 \]

\[ \frac{2h}{\sqrt{3}} = 100 \quad\Longrightarrow\quad h = 50\sqrt{3} \approx 86.6 \text{ m} \]

The pairing \( 30^{\circ}/60^{\circ} \) is convenient because \( \sqrt{3} - \tfrac{1}{\sqrt{3}} \) collapses to \( \tfrac{2}{\sqrt{3}} \).

Answer: B

The temple stands on one bank, so the river's width is the gap between the two sight lines - the difference of the horizontal distances.

\[ d_{45} = \frac{100}{\tan 45^{\circ}} = 100 \text{ m} \]

\[ d_{30} = \frac{100}{\tan 30^{\circ}} = 100\sqrt{3} \text{ m} \]

\[ \text{width} = 100\sqrt{3} - 100 = 100(\sqrt{3} - 1) \text{ m} \]

That is roughly 73.2 m. The nearer bank gives the steeper \( 45^{\circ} \) depression, the far bank the shallower \( 30^{\circ} \).

Answer: C

Let the tower be \( h \). The two observation points give:

\[ d_{30} = h\sqrt{3}, \qquad d_{45} = h \]

\[ h\sqrt{3} - h = 40 \quad\Longrightarrow\quad h(\sqrt{3}-1) = 40 \]

Rationalising:

\[ h = \frac{40}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{40(\sqrt{3}+1)}{2} = 20(\sqrt{3}+1) \text{ m} \]

Numerically about 54.6 m. Multiplying the numerator and denominator by the conjugate is what turns the surd denominator into the plain 2.

Answer: A

The shadow is the base of the right triangle and the building is the side opposite the sun's \( 60^{\circ} \) elevation.

\[ \tan 60^{\circ} = \frac{h}{20} \]

\[ h = 20\sqrt{3} = 20 \times 1.732 = 34.64 \text{ m} \]

With the sun above \( 45^{\circ} \) the building must be taller than its shadow, so any answer below 20 m can be discarded immediately.

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.