Heights & Distances Questions and Answers

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: B

Let the tower be \( T \) m high and \( d \) m away. Two sightings:

\[ \tan b = \frac{T}{d} = \frac{4}{5}, \qquad \tan a = \frac{T - 230}{d} = \frac{5}{12} \]

Subtracting eliminates \( T \):

\[ \frac{230}{d} = \frac{4}{5} - \frac{5}{12} = \frac{48 - 25}{60} = \frac{23}{60} \]

\[ d = 230 \times \frac{60}{23} = 600 \text{ m} \]

\[ T = \frac{4}{5} \times 600 = 480 \text{ m}, \qquad T - 230 = 250 \text{ m} \]

The line joining the two tops is the hypotenuse of a right triangle with legs \( d \) and \( T - 230 \):

\[ \sqrt{600^{2} + 250^{2}} = \sqrt{360000 + 62500} = \sqrt{422500} = 650 \text{ m} \]

600 m is the horizontal distance only; the question asks for the slant distance between the tops.

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

Angles of depression from the top equal the angles of elevation from the car, so for a tower of height \( h \):

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

In 8 minutes the car covers \( h\sqrt{3} - h \); the distance left is \( h \). At constant speed time is proportional to distance:

\[ t = 8 \times \frac{h}{h(\sqrt{3}-1)} = \frac{8}{\sqrt{3}-1} = \frac{8(\sqrt{3}+1)}{2} = 4(\sqrt{3}+1) \]

\[ t = 4 \times 2.732 = 10.93 \text{ min} \approx 10 \text{ min } 57 \text{ s} \]

The height \( h \) cancels, so the tower's size is irrelevant - which is why the question can omit it.

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: 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: 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: 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: 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: A

Fig.
Let AB be the hill whose foot is B and D and C are two kilometer stones. 
Therefore DC = 1 km = 1000 m,
 Let AB = h and BC = x.
In right angled Δ ABC,
tan 45º = AB/BC
=> 1 = h/x
=> x = h                                                       ------------- (1) 
In right angled Δ ABC,
 tan 30º = AB/BD
=> 1/√3 = h/(x + 1000) 
=> x + 1000 = h√3                                            ----------------(2) 
using (1) and (2) we get
x + 1000 = x√3
 => x(√3 - 1) = 1000 
Or, x = [1000/(√3 - 1)] × (√3 + 1)/(√3 + 1) 
Or, x = 1000 (√3 + 1)/2
= 500(√3 + 1)
= 500 × 2.732 
= 1366 m or 1.366 km. 
Therefore first km stone is 1.366 km and second km stone is 2.366 km from the foot of the hill.