Heights & Distances — Chapter Test

The angle of elevation of the top of a tower from a point A (on the ground) is 30°. On walking 50 m towards the tower, the angle of elevation is found to be 60°. Calculate :

(i) the height of the tower (correct to one decimal place)

(ii) the distance of the tower from A.

Answer

Consider the below figure:

(i) Let after moving 50 m towards tower from point A, the person reaches point D and height of tower be h meters.

From figure,

AD = 50 m, AB = AD + DB = (50 + DB) m

Considering right angled triangle △ABC,

$\Rightarrow \text{tan 30°} = \dfrac{BC}{AB} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{h}{50 + DB} \\[1em] \Rightarrow 50 + DB = h\sqrt{3} \text{ .......( Eq 1)}$

Considering right angled triangle △BCD,

$\Rightarrow \text{tan 60°} = \dfrac{BC}{DB} \\[1em] \Rightarrow \sqrt{3} = \dfrac{h}{DB} \\[1em] \Rightarrow h = DB\sqrt{3} \text{ .......( Eq 2)}$

Putting value of h from Eq 2 in Eq 1 we get,

$\Rightarrow 50 + DB = DB\sqrt{3} \times \sqrt{3} \\[1em] \Rightarrow 50 + DB = 3DB \\[1em] \Rightarrow 2DB = 50 \\[1em] \Rightarrow DB = 25 \\[1em] \therefore h = DB\sqrt{3} = 25 \times 1.732 = 43.3 \text{ m}$

Hence, the height of tower is 43.3 m.

(ii) From figure,

Distance of tower from A (AB) = AD + DB = 50 + 25 = 75 m.

Hence, the distance of tower from A is 75 m.

An aeroplane 3000 m high, passes vertically above another aeroplane at an instant when the angles of elevation of the two aeroplanes from the same point on the ground are 60° and 45° respectively. Find the vertical distance between the two planes.

Answer

Let the plane 3000 m high be at point B and plane below it be at point D.

From figure,

Considering right angled triangle △ABC,

$\Rightarrow \text{tan 60°} = \dfrac{BC}{AC} \\[1em] \Rightarrow \sqrt{3} = \dfrac{3000}{AC} \\[1em] \Rightarrow AC = \dfrac{3000}{\sqrt{3}} \\[1em] \Rightarrow AC = 1732 \text{ m}.$

Considering right angled triangle △ADC,

$\Rightarrow \text{tan 45°} = \dfrac{DC}{AC} \\[1em] \Rightarrow 1 = \dfrac{DC}{1732} \\[1em] \Rightarrow DC = 1732 \text{ m}.$

Distance between two planes (BD) = BC - DC = 3000 - 1732 = 1268 m.

Hence, the vertical distance between two planes is 1268 m.

A 7 m long flagstaff is fixed on the top of a tower. From a point on the ground, the angles of elevation of the top and bottom of the flagstaff are 45° and 36° respectively. Find the height of the tower correct to one place of decimal.

Answer

Let CD be the tower of height h meters and BD the flagstaff.

A be point on the ground from where the angles of elevation of the top and bottom of the flagstaff are 45° and 36° respectively.

From figure,

BC = BD + DC = (7 + h) meters.

Considering right angled triangle △ABC,

$\Rightarrow \text{tan 45°} = \dfrac{BC}{AC} \\[1em] \Rightarrow 1 = \dfrac{7 + h}{AC} \\[1em] \Rightarrow AC = h + 7 \text{ ......(Eq 1)}.$

Considering right angled triangle △ADC,

$\Rightarrow \text{tan 36°} = \dfrac{DC}{AC} \\[1em] \Rightarrow 0.7265 = \dfrac{h}{AC} \\[1em] \Rightarrow AC = \dfrac{h}{0.7265}$

Putting value of AC in Eq 1 we get,

$\Rightarrow \dfrac{h}{0.7265} = h + 7 \\[1em] \Rightarrow h = 0.7265(h + 7) \\[1em] \Rightarrow h = 0.7265h + 5.0855 \\[1em] \Rightarrow h - 0.7265\text{ h} = 5.0855 \\[1em] \Rightarrow 0.2735\text{ h} = 5.0855 \\[1em] \Rightarrow h = \dfrac{5.0855}{0.2735} \\[1em] \Rightarrow h = 18.6 \text{ m}.$

Hence, the height of tower is 18.6 m.

A boy, 1.6 m tall, is 20 m away from a tower and observes that the angle of elevation of the top of the tower is 60°. Find the height of the tower.

Answer

Let AD be man and BC be tower of height h meters.

From figure,

ABED is a rectangle.

BE = AD = 1.6 m
DE = AB = 20 m.

CE = BC - BE = (h - 1.6) m.

Considering right angled triangle △DCE,

$\Rightarrow \text{tan 60°} = \dfrac{CE}{DE} \\[1em] \Rightarrow \sqrt{3} = \dfrac{h - 1.6}{20} \\[1em] \Rightarrow h - 1.6 = 20\sqrt{3} \\[1em] \Rightarrow h - 1.6 = 34.64 \\[1em] \Rightarrow h = 36.24 \text{ m}.$

Hence, the height of tower is 36.24 m.

A boy 1.54 m tall can just see the sun over a wall 3.64 m high which is 2.1 m away from him. Find the angle of elevation of the sun.

Answer

Let AD be man, BC be tower and θ be the angle of elevation.

From figure,

ABED is a rectangle.

BE = AD = 1.54 m
DE = AB = 2.1 m.

CE = BC - BE = (3.64 - 1.54) = 2.1 m.

Considering right angled triangle △DCE,

$\Rightarrow \text{tan θ} = \dfrac{CE}{DE} \\[1em] \Rightarrow \text{tan θ} = \dfrac{2.1}{2.1} \\[1em] \Rightarrow \text{tan θ} = 1 \\[1em] \Rightarrow \text{tan θ} = \text{tan 45°} \\[1em] \Rightarrow θ = 45°$

Hence, the angle of elevation is 45°.

An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60°. After 10 seconds, its elevation is observed to be 30°. Find the speed of the aeroplane in km/h.

Answer

Let initially aeroplane be at point B and after 10 seconds it is at point C.

Since, aeroplane is flying horizontally 1 km above the ground so, BE = CD = 1 km.

Considering right angled triangle △ABE,

$\Rightarrow \text{tan 60°} = \dfrac{BE}{AE} \\[1em] \Rightarrow \sqrt{3} = \dfrac{1}{AE} \\[1em] \Rightarrow AE = \dfrac{1}{\sqrt{3}} \\[1em] \Rightarrow AE = 0.577 \text{ km}.$

From figure,

AD = AE + ED = (0.577 + ED) km.

Considering right angled triangle △ACD,

$\Rightarrow \text{tan 30°} = \dfrac{CD}{AD} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{1}{0.577 + ED} \\[1em] \Rightarrow 0.577 + ED = \sqrt{3} \\[1em] \Rightarrow ED = 1.732 - 0.577 \\[1em] \Rightarrow ED = 1.155 \text{ km}.$

Aeroplane covers 1.155 km in 10 seconds.

Time = 10 seconds = $\dfrac{10}{60 \times 60} = \dfrac{10}{3600} = \dfrac{1}{360}$ hours.

Speed = $\dfrac{\text{Distance}}{\text{Time}} = \dfrac{1.155}{\dfrac{1}{360}} = 1.155 \times 360 =$ 415.66 km/h.

Hence, the speed of aeroplane is 415.66 km/h.

A man on the deck of a ship is 16 m above the water level. He observes that the angle of elevation of the top of a cliff is 45° and the angle of depression of the base is 30°. Calculate the distance of the cliff from the ship and the height of the cliff.

Answer

Let A be the man on the deck of the ship B and CE is the cliff.

AB = 16 m and angle of elevation of the top of cliff is 45° and angle of depression of base of cliff is 30°.

Let CE = h, AD = x, then
CD = h - 16, AD = BE = x.

Now in right angled triangle △CAD,

$\text{tan 45°} = \dfrac{CD}{AD} \\[1em] 1 = \dfrac{h - 16}{x} \\[1em] x = h - 16 \text{ ......(i)}$

Again in right angled triangle △ADE,

$\text{tan 30°} = \dfrac{DE}{AD} = \dfrac{16}{x} \\[1em] \dfrac{1}{\sqrt{3}} = \dfrac{16}{x} \\[1em] x = 16\sqrt{3} \\[1em] x = 27.71 \text{ m} \text{ ......(ii)}$

From (i) and (ii),

$\Rightarrow h - 16 = 27.71 \\[1em] \Rightarrow h = 27.71 + 16 \\[1em] \Rightarrow h = 43.71 \text{ m}$

Hence, the distance of cliff from the ship is 27.71 m and height of cliff is 43.71 m.

There is a small island in between a river 100 meters wide. A tall tree stands on the island. P and Q are points directly opposite to each other on the two banks, and in line with the tree. If the angles of elevation of the top of the tree from P and Q are 30° and 45° respectively, find the height of the tree.

Answer

Let XY be tree of h meters.

From figure,

Considering right angled triangle △XQY,

$\Rightarrow \text{tan 45°} = \dfrac{XY}{YQ} \\[1em] \Rightarrow 1 = \dfrac{h}{YQ} \\[1em] \Rightarrow YQ = h \text{ ....(Eq 1)}$

Considering right angled triangle △XPY,

$\Rightarrow \text{tan 30°} = \dfrac{XY}{PY} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{h}{100 - YQ} \\[1em] \Rightarrow 100 - YQ = h\sqrt{3}$

Putting value of YQ from Eq 1 in above equation,

$\Rightarrow 100 - h = 1.732h \\[1em] \Rightarrow 1.732h + h = 100 \\[1em] \Rightarrow 2.732h = 100 \\[1em] \Rightarrow h = \dfrac{100}{2.732} = 36.6 \text{ m}.$

Hence, the height of tree is 36.6 m.

A man standing on the deck of the ship which is 20 m above the sea level, observes the angle of elevation of a bird as 30° and the angle of depression of its reflection in the sea as 60°. Find the height of the bird.

Answer

Let P be the man standing on the deck of the ship which is 20 m above sea level and B is the bird.

Let bird be flying above h meters from deck of the ship.

∴ BC = h

Let the shadow be created at point R, the shadow will be created vertically opposite at same distance as bird is from ship.

∴ AR = AB = AC + BC = h + 20.

From figure,

In right angled triangle △PCB,

$\Rightarrow \text{tan 30°} = \dfrac{BC}{PC} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{h}{x} \\[1em] \Rightarrow x = h\sqrt{3} ......(i)$

In right angled triangle △PCR,

$\Rightarrow \text{tan 60°} = \dfrac{CR}{CP} \\[1em] \Rightarrow \sqrt{3} = \dfrac{h + 40}{x} \\[1em] \Rightarrow \dfrac{h + 40}{h\sqrt{3}} = \sqrt{3} \quad \text{ [From (i)]}\\[1em] \Rightarrow h + 40 = \sqrt{3} \times \sqrt{3}h \\[1em] \Rightarrow h + 40 = 3h \\[1em] \Rightarrow 3h - h = 40 \\[1em] \Rightarrow 2h = 40 \\[1em] \Rightarrow h = 20.$

From sea level, height of bird (AB) = h + 20 = 20 + 20 = 40 m.

Hence, the height of bird from sea level is 40 m.

A vertical tower standing on a horizontal plane is surmounted by a vertical flagstaff. At a point 100 m away from the foot of the tower, the angle of elevation of the top and bottom of the flagstaff are 54° and 42° respectively. Find the height of the flagstaff. Give your answer correct to nearest metre.

Answer

In △APB,

⇒ tan 42° = $\dfrac{AB}{AP}$

⇒ 0.9004 = $\dfrac{AB}{100}$

⇒ AB = 0.9004 × 100 = 90.04 m

In △APF,

⇒ tan 54° = $\dfrac{AF}{AP}$

⇒ 1.3764 = $\dfrac{AF}{100}$

⇒ AF = 1.3764 × 100 = 137.64 m

From figure,

BF = AF - AB = 137.64 - 90.04 = 47.60 ≈ 48 m.

Hence, height of flagstaff = 48 m.