Heights & Distances Questions and Answers

Answer: D

Fig.
Let AD be the tree of height h,
In ΔADC,
 tan60º = h/CD 
Or, √3 = h/CD 
Or, CD = h/√3 
In ΔADB,
tan30º = h/BD 
Or, 1/√3 = h/BD 
Or, BD = h√3 
BD - CD = 50 
h√3 - h/√3 = 50 
(3h - h)/√3 = 50 
2h = 50√3
Height of the tree,
h = 50√3/2
   = 25√3 25×1.732
   = 43.3 m.
Width of the river,
CD = h/√3
      = 25√3/√3
      = 25 m. 

Answer: B

F is the foot of the tower, A and B are on the same straight line from it, with \( AF = 200 \) m and \( AB = 80 \) m, so \( BF = 120 \) m.

(a) Height of the tower. From A:

\[ \tan x^{\circ} = \frac{TF}{AF} = \frac{2}{5} \quad\Longrightarrow\quad TF = 200 \times \frac{2}{5} = 80 \text{ m} \]

(b) The angle y. From B, the same 80 m tower is only 120 m away:

\[ \tan y^{\circ} = \frac{80}{120} = \frac{2}{3} = 0.6667 \]

\[ y = \tan^{-1}(0.6667) = 33.69^{\circ} \approx 34^{\circ} \]

Moving 80 m closer raises the angle from about \( 21.8^{\circ} \) to \( 34^{\circ} \) - the elevation grows faster the nearer you get.

Answer: A

Let L SPR = θ then L SQR = 90º - θ and RS = h, 
In Δ SPR, tan θ = h/a Or, h = a tan θ --------------------------------- (i) 
In ΔSQR, tan (90º - θ) = h/b Or, h = b tan (90º - θ) = h cot θ ----- (ii) 
Multiplying (i) and (ii) we get, h2 = a tan θ b cot θ = a b 
Or, h = √(a b). [Proved.] 

Answer: C

Work out the horizontal distance first, using the depression of the tower's foot: the window is 10 m above ground, so that 10 m is the opposite side.

\[ \tan y^{\circ} = \frac{10}{d} = \frac{1}{4} \quad\Longrightarrow\quad d = 40 \text{ m} \]

Now the elevation of the tower's top gives the part above window level:

\[ \tan x^{\circ} = \frac{h}{40} = \frac{5}{2} \quad\Longrightarrow\quad h = 100 \text{ m} \]

The full tower runs from the ground to C, so add the window's own height:

\[ CD = 100 + 10 = 110 \text{ m} \]

Forgetting the 10 m leaves 100 m, which is why that value is not offered - the near miss on the list is 70 m, from using \( \tan y \) the wrong way up.

Answer: B

The shadow lengthens as the sun drops from \( 45^{\circ} \) to \( 30^{\circ} \).

\[ s_{45} = \frac{h}{\tan 45^{\circ}} = h, \qquad s_{30} = \frac{h}{\tan 30^{\circ}} = h\sqrt{3} \]

\[ h\sqrt{3} - h = 10 \quad\Longrightarrow\quad h(\sqrt{3}-1) = 10 \]

\[ h = \frac{10}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{10(\sqrt{3}+1)}{2} = 5(\sqrt{3}+1) \]

\[ h = 5 \times 2.732 = 13.66 \text{ m} \]

The 10 m is the increase in shadow, not a shadow length in its own right.

Answer: B

Let AB be the tower and C and D the poles. AB = 70 m. 
In Δ ABC, cot 50º = BC/AB 
Or, cot (90º - 40º) = BC/70 
Or, tan 40º = BC/70 
Or, 0.8391 = BC/70 
Or, BC = 0.8391 × 70 = 58.74 m. 
In Δ ABD, cot 25º = BD/AB 
Or, cot (90º - 65º) = BD/70 
Or, tan 65º = BD/70 
Or, 2.114451 = BD/70 
Or, BD = 2.114451 × 70 = 148.01 m. 
Hence, distance between people = CD = BD - BC 
= 148.01 - 58.74 = 89.27 m.

Answer: C

Correct answer: \(30\sqrt{3}\) m

The height and the base distance are the opposite and adjacent sides of the angle of elevation.

\[\tan 60^\circ = \frac{h}{30} \quad\Longrightarrow\quad h = 30\tan 60^\circ = 30\sqrt{3} \approx 51.96 \text{ m}\]

Answer: D

The ladder is the hypotenuse and the 4.6 m ground gap is adjacent to the \( 60^{\circ} \) elevation, so cosine is the ratio.

\[ \cos 60^{\circ} = \frac{4.6}{L} \quad\Longrightarrow\quad \frac{1}{2} = \frac{4.6}{L} \]

\[ L = 2 \times 4.6 = 9.2 \text{ m} \]

Since \( \cos 60^{\circ} = \tfrac{1}{2} \), the ladder is exactly twice the foot distance. Using tangent instead returns \( 4.6\sqrt{3} = 7.97 \) m, which is the height reached on the wall - hence the 7.8 m distractor.

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 ships lie on opposite sides, so the lighthouse foot is between them and the two horizontal distances add.

\[ d_{30} = \frac{100}{\tan 30^{\circ}} = 100\sqrt{3} = 173.2 \text{ m} \]

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

\[ \text{distance apart} = 173.2 + 100 = 273.2 \approx 273 \text{ m} \]

Had the ships been on the same side the answer would be the difference, 73 m. Answering 173 m counts only the \( 30^{\circ} \) ship.