Heights & Distances — Test Yourself

If CD = 10 m, the length of AB is :

1. $\dfrac{40}{\sqrt{3}}$ m

2. $\dfrac{20}{\sqrt{3}}$ m

3. 100 m

4. 30 m

Answer

We know that,

tan θ = $\dfrac{\text{Perpendicular}}{\text{Base}}$

From figure,

In △ BCD,

⇒ tan 30° = $\dfrac{BD}{CD}$

⇒ $\dfrac{1}{\sqrt{3}} = \dfrac{BD}{10}$

⇒ BD = $\dfrac{10}{\sqrt{3}}$ m.

In △ ACD,

⇒ tan 60° = $\dfrac{AD}{CD}$

⇒ $\sqrt{3} = \dfrac{AD}{10}$

⇒ AD = $10\sqrt{3}$ cm.

From figure,

AB = AD + BD = $10\sqrt{3} + \dfrac{10}{\sqrt{3}}$

$= \dfrac{10\sqrt{3} \times \sqrt{3} + 10}{\sqrt{3}} = \dfrac{30 + 10}{\sqrt{3}} = \dfrac{40}{\sqrt{3}}$ m.

Hence, Option 1 is the correct option.

If tan x° = $\dfrac{5}{2}$, the length of CB is :

1. 0.4 m

2. 40 m

3. 50 m

4. 80 m

Answer

We know that,

tan θ = $\dfrac{\text{Perpendicular}}{\text{Base}}$

From figure,

In △ ABC,

⇒ tan x° = $\dfrac{AB}{CB}$

⇒ $\dfrac{5}{2} = \dfrac{100}{CB}$

⇒ CB = $\dfrac{2 \times 100}{5} = \dfrac{200}{5}$ = 40 m.

Hence, Option 2 is the correct option.

The length of AC is :

1. $(60 - 20\sqrt{3})$ m

2. $60\sqrt{3}$ m

3. $(60 + 20\sqrt{3})$ m

4. $20\sqrt{3}$ m

Answer

In rectangle BCDE,

Opposite sides of rectangle are equal.

∴ BE = DC = 60 m

We know that,

tan θ = $\dfrac{\text{Perpendicular}}{\text{Base}}$

From figure,

In △ ABE,

⇒ tan 30° = $\dfrac{AB}{BE}$

⇒ $\dfrac{1}{\sqrt{3}} = \dfrac{AB}{60}$

⇒ AB = $\dfrac{60}{\sqrt{3}} = 20\sqrt{3}$ m.

In △ EBC,

⇒ tan 45° = $\dfrac{BC}{BE}$

⇒ $1 = \dfrac{BC}{60}$

⇒ BC = 60 m.

From figure,

AC = AB + BC = $(20\sqrt{3} + 60)$ m.

Hence, Option 3 is the correct option.

BCDE is a square with side 90 cm and ∠F = 45°. The length of AF is :

1. $60\sqrt{2}$ cm

2. $90\sqrt{2}$ cm

3. $120\sqrt{2}$ cm

4. $180\sqrt{2}$ cm

Answer

Since, BCDE is a square.

∴ BE = CD = 90 cm.

We know that,

tan θ = $\dfrac{\text{Perpendicular}}{\text{Base}}$

From figure,

∠AEB = ∠EFD = 45° (Corresponding angles are equal)

In △ ABE,

⇒ tan 45° = $\dfrac{AB}{BE}$

⇒ $1 = \dfrac{AB}{90}$

⇒ AB = 90 cm.

From figure,

AC = AB + BC = 90 + 90 = 180 cm.

We know that,

sin θ = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$

In △ ACF,

⇒ sin 45° = $\dfrac{AC}{AF}$

⇒ $\dfrac{1}{\sqrt{2}} = \dfrac{180}{AF}$

⇒ AF = $180\sqrt{2}$ cm.

Hence, Option 4 is the correct option.

The length of DC is :

1. $45\sqrt{3}$ m

2. $(45 - \sqrt{3})$ m

3. $45(\sqrt{3} - 1)$ m

4. $45(\sqrt{3} + 1)$ m

Answer

We know that,

tan θ = $\dfrac{\text{Perpendicular}}{\text{Base}}$

From figure,

In △ ABC,

⇒ tan 45° = $\dfrac{AB}{BC}$

⇒ $1 = \dfrac{AB}{45}$

⇒ AB = 45 m.

In △ ABD,

⇒ tan 30° = $\dfrac{AB}{BD}$

⇒ $\dfrac{1}{\sqrt{3}} = \dfrac{45}{BD}$

⇒ BD = $45\sqrt{3}$ m.

From figure,

DC = BD - BC = $45\sqrt{3} - 45 = 45(\sqrt{3} - 1)$ m.

Hence, Option 3 is the correct option.

The ratio of the length of a vertical pole and length of its shadow on the horizontal surface is 3 : $\sqrt{3}$

Assertion(A): The angle of elevation of the sun is 60°.

Reason(R): If angle of elevation of the sun is θ, tan θ = $\dfrac{3}{\sqrt{3}} = \sqrt{3}$ = tan 60°.

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.

Answer

Given, the ratio of the height of a vertical pole to the length of its shadow is 3 : $\sqrt{3}$.

Let AB be the pole and BC be the shadow and angle of elevation of Sun be θ.

$\Rightarrow \dfrac{\text{Height of a vertical pole}}{\text{Height of shadow}} = \dfrac{3}{\sqrt{3}}\\[1em] \Rightarrow \dfrac{\text{AB}}{\text{BC}} = \sqrt{3}\\[1em] \Rightarrow \text{tan θ} = \sqrt{3}\\[1em] \Rightarrow \text{tan θ} = \text{tan 60°} \\[1em] \Rightarrow θ = 60°$

∴ Both A and R are true and R is correct reason for A.

Hence, option 3 is the correct option.

The length of the ladder placed against a vertical wall is twice the distance between the foot of the ladder and the wall.

Assertion(A): The angle that the ladder makes with the wall is 60°.

Reason(R): If ladder makes angle θ with the wall then sin θ = $\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{x}{2x} = \dfrac{1}{2}$.

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.

Answer

Let AB be the wall and AC be the ladder. Let distance between foot of ladder and wall be x.

So, length of ladder (AC) = 2x.

According to the Pythagoras theorem :

⇒ Hypotenuse² = Base² + Height²

⇒ AC² = BC² + AB²

⇒ (2x)² = x² + AB²

⇒ 4x² = x² + AB²

⇒ AB² = 4x² - x²

⇒ AB² = 3x²

⇒ AB = $\sqrt{3x^2}$

⇒ AB = $\sqrt{3}x$.

Let angle between ladder and wall be θ.

We know that,

$\Rightarrow \text{sin θ} = \dfrac{\text{Perpendicular side}}{\text{Hypotenuse}} \\[1em] \Rightarrow \text{sin θ} = \dfrac{BC}{AC} \\[1em] \Rightarrow \text{sin θ} = \dfrac{x}{2x} \\[1em] \Rightarrow \text{sin θ} = \dfrac{1}{2} \\[1em]$

⇒ sin θ = sin 30°

⇒ θ = 30°.

Assertion (A) is false.

If ladder makes angle θ with the wall then sin θ = $\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{x}{2x} = \dfrac{1}{2}$.

Here base refers side opposite to angle θ i.e. BC.

Reason (R) is true.

Hence, option 2 is the correct option.

For the following figure, tan A = $1\dfrac{2}{3}$ and tan B = $\dfrac{1}{3}$

Statement (1): x = 36 cm

Statement (2): tan A = $\dfrac{5}{3} = \dfrac{h}{9}$ ⇒ h = 15

tan B = $\dfrac{h}{x + 9} \Rightarrow \dfrac{1}{3} = \dfrac{15}{x + 9}$

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Given,

tan A = $1\dfrac{2}{3} = \dfrac{5}{3}$ and tan B = $\dfrac{1}{3}$.

From figure,

$\Rightarrow \text{tan A} = \dfrac{h}{9} \\[1em] \Rightarrow \dfrac{5}{3} = \dfrac{h}{9} \\[1em] \Rightarrow h = \dfrac{5 \times 9}{3} \\[1em] \Rightarrow h = 5 \times 3 \\[1em] \Rightarrow h = 15\text{ cm} \\[1em] \Rightarrow \text{tan B} = \dfrac{h}{x + 9}\\[1em] \Rightarrow \dfrac{1}{3} = \dfrac{15}{x + 9}\\[1em] \Rightarrow x + 9 = 15 \times 3 \Rightarrow x + 9 = 45 \Rightarrow x = 45 - 9 \Rightarrow x = 36 \text{ cm}.$

∴ Both the statements are true.

Hence, option 1 is the correct option.

CD = 100 m, ∠ADB = 15° and ∠BDC = 45°

Statement (1): AB = $\dfrac{\text{tan 60° - tan 45°}}{100}$

Statement (2): AB = (100 tan 60° - 100 tan 45°) m.

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

From figure,

In triangle BDC,

$\Rightarrow \text{tan 45°} = \dfrac{BC}{DC}\\[1em] \Rightarrow \text{tan 45°} = \dfrac{BC}{100}\\[1em] \Rightarrow BC = 100\text{ tan 45°}\\[1em] \text{In triangle ADC,}\\[1em] \Rightarrow \text{tan 60°} = \dfrac{AC}{DC}\\[1em] \Rightarrow \text{tan 60°} = \dfrac{AB + BC}{DC}\\[1em] \Rightarrow \text{tan 60°} = \dfrac{AB + 100\text{ tan 45°}}{100}\\[1em] \Rightarrow 100\text{ tan 60°} = AB + 100\text{ tan 45°}\\[1em] \Rightarrow AB = 100\text{ tan 60°} - 100\text{ tan 45°}$

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Find AD.

(i)

(ii)

Answer

(i) From figure,

BE = CD = 20 m and DE = CB = 5 m.

In △ABE,

$\text{tan 32°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 0.6249 = \dfrac{AE}{BE} \\[1em] \Rightarrow AE = 0.6249 \times BE \\[1em] \Rightarrow AE = 0.6249 \times 20 \\[1em] \Rightarrow AE = 12.498 \text{ m}.$

AD = AE + DE = 12.498 + 5 = 17.498 ≈ 17.5 m.

Hence, AD = 17.5 meters.

(ii) We know that,

An exterior angle is equal to the sum of two opposite interior angles.

∴ ∠ACD = ∠ABC + ∠BAC

Also, ∠ABC = ∠BAC (As, angles opposite to equal sides are equal)

∴ ∠ACD = 2∠ABC

⇒ 2∠ABC = 48°

⇒ ∠ABC = 24°.

In △ABD,

$\text{sin 24°} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] \Rightarrow 0.4067 = \dfrac{AD}{AB} \\[1em] \Rightarrow AD = 0.4067 \times AB \\[1em] \Rightarrow AD = 0.4067 \times 30 \\[1em] \Rightarrow AD = 12.20 \text{ m}.$

Hence, AD = 12.20 meters.

In the following diagram, AB is a floor-board; PQRS is a cubical box with each edge = 1 m and ∠B = 60°. Calculate the length of the board AB.

Answer

In ∆PSB,

$\text{sin 60°} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] \Rightarrow \dfrac{\sqrt{3}}{2} = \dfrac{PS}{PB} \\[1em] \Rightarrow PB = \dfrac{2PS}{\sqrt{3}} \\[1em] \Rightarrow PB = \dfrac{2 \times 1}{1.732} = 1.155 \text{ m}.$

In ∆APQ,

∠APQ = ∠ABR = 60° (Corresponding angles are equal.)

$\text{cos 60°} = \dfrac{\text{Base}}{\text{Hypotenuse}} \\[1em] \Rightarrow \dfrac{1}{2} = \dfrac{PQ}{AP} \\[1em] \Rightarrow AP = 2PQ \\[1em] \Rightarrow AP = 2 \text{ m}.$

From figure,

AB = AP + PB = 2 + 1.155 = 3.155 m.

Hence, AB = 3.155 meters.

Calculate BC.

Answer

In ∆ABD,

$\text{tan 35°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 0.7002 = \dfrac{AD}{BD} \\[1em] \Rightarrow BD = \dfrac{AD}{0.7002} \\[1em] \Rightarrow BD = \dfrac{20}{0.7002} \\[1em] \Rightarrow BD = 28.563 \text{ m}.$

In ∆ACD,

$\text{tan 42°} = \dfrac{\text{Perpendicular}}{Base} \\[1em] \Rightarrow 0.9004 = \dfrac{CD}{AD} \\[1em] \Rightarrow CD = AD \times 0.9004 \\[1em] \Rightarrow CD = 20 \times 0.9004 \\[1em] \Rightarrow CD = 18.008 \text{ m}.$

From figure,

BC = BD - CD = 28.563 - 18.008 = 10.55 meters.

Hence, BC = 10.55 meters.

Calculate AB.

Answer

In △ACD,

$\text{cos 30°} = \dfrac{\text{Base}}{\text{Hypotenuse}} \\[1em] \Rightarrow \dfrac{\sqrt{3}}{2} = \dfrac{AD}{CD} \\[1em] \Rightarrow AD = \dfrac{\sqrt{3}}{2} \times CD \\[1em] \Rightarrow AD = \dfrac{\sqrt{3}}{2} \times 6 \\[1em] \Rightarrow AD = 3\sqrt{3} \\[1em] \Rightarrow AD = 5.196 \text{ m}.$

In △BDE,

$\text{sin 47°} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] \Rightarrow 0.73 = \dfrac{BD}{DE} \\[1em] \Rightarrow BD = 0.73 \times DE \\[1em] \Rightarrow BD = 0.73 \times 5 = 3.65 \text{ m}.$

From figure,

AB = AD + BD = 5.196 + 3.65 = 8.846 = 8.85 meters.

Hence, AB = 8.85 meters.

The radius of a circle is given as 15 cm and chord AB subtends an angle of 131° at the centre C of the circle. Using trigonometry, calculate :

(i) the length of AB;

(ii) the distance of AB from the centre C.

Answer

Given,

CA = CB = 15 cm and ∠ACB = 131°.

Construct a perpendicular CP from center C to the chord AB.

We know that perpendicular form center to the chord bisects the chord.

Then, CP bisects AB.

In △ACP and △BCP,

∠APC = ∠BPC = 90°

CP = CP [∵ Common Side]

AP = PB [∵ CP bisects AB]

∴ △ACP ≅ △BCP by SAS axiom.

∴ ∠ACP = ∠BCP = $\dfrac{131°}{2}$ = 65.5° [By C.P.C.T.]

In △ACP,

$\text{sin 65.5°} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] \Rightarrow 0.91 = \dfrac{AP}{AC} \\[1em] \Rightarrow AP = 0.91 \times AC \\[1em] \Rightarrow AP = 0.91 \times 15 = 13.65 \text{ cm}.$

(i) From figure,

AB = AP + PB = 2AP

= 2 × 13.65

= 27.30 cm

Hence, AB = 27.30 cm.

(ii) In △ACP,

$\text{cos 65.5°} = \dfrac{\text{Base}}{\text{Hypotenuse}} \\[1em] \Rightarrow 0.415 = \dfrac{CP}{AC} \\[1em] \Rightarrow CP = 0.415 \times 15 \\[1em] \Rightarrow CP = 6.225 \text{ cm}.$

Hence, CP = 6.225 cm.

At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is $\dfrac{5}{12}$. On walking 192 meters towards the tower; the tangent of the angle is found to be $\dfrac{3}{4}$. Find the height of the tower.

Answer

Let's assume AB to be the vertical tower and C and D be the two points such that CD = 192 m.

Let ∠ACB = θ and ∠ADB = α

Given,

$\Rightarrow \text{tan θ} = \dfrac{5}{12} \\[1em] \Rightarrow \dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{5}{12} \\[1em] \Rightarrow \dfrac{AB}{BC} = \dfrac{5}{12} \\[1em] \Rightarrow AB = \dfrac{5}{12}BC .........(1)$

Also given,

$\Rightarrow \text{tan α} = \dfrac{3}{4} \\[1em] \Rightarrow \dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{3}{4} \\[1em] \Rightarrow \dfrac{AB}{BD} = \dfrac{3}{4} \\[1em] \Rightarrow AB = \dfrac{3}{4}BD .........(2)$

From equations (1) and (2), we get :

$\Rightarrow \dfrac{5}{12}BC = \dfrac{3}{4}BD \\[1em] \Rightarrow \dfrac{BC}{BD} = \dfrac{3 \times 12}{4 \times 5} \\[1em] \Rightarrow \dfrac{BD + CD}{BD} = \dfrac{36}{20} \\[1em] \Rightarrow \dfrac{BD + CD}{BD} = \dfrac{9}{5} \\[1em] \Rightarrow 5(BD + CD) = 9BD \\[1em] \Rightarrow 5BD + 5CD = 9BD \\[1em] \Rightarrow 9BD - 5BD = 5CD \\[1em] \Rightarrow 4BD = 5CD \\[1em] \Rightarrow BD = \dfrac{5 \times 192}{4} \\[1em] \Rightarrow BD = \dfrac{960}{4} \\[1em] \Rightarrow BD = 240 \text{ m}.$

BC = BD + DC = 240 + 192 = 432 m.

From equation (1),

AB = $\dfrac{5}{12} \times BC = \dfrac{5}{12} \times 432$

= 5 × 36

= 180 m.

Hence, the height of the tower is 180 m.

A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h meter. At a point on the plane, the angle of elevation of the bottom of the flagstaff is α and at the top of the flagstaff is β. Prove that the height of the tower is $\dfrac{\text{h tan α}}{\text{tan β} - \text{tan α}}$.

Answer

Let AB be the tower of height x meters, surmounted by a vertical flagstaff AD of height h meters (given). Let C be a point on the plane such that ∠ACB = α, ∠DCB = β and AD = h.

In ∆ABC,

$\Rightarrow \text{tan α} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan α}= \dfrac{AB}{BC} \\[1em] \Rightarrow BC = \dfrac{AB}{\text{tan α}} \\[1em] \Rightarrow BC = \dfrac{x}{\text{tan α}} ..........(1)$

In ∆DBC,

$\Rightarrow \text{tan β} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan β} = \dfrac{BD}{BC} \\[1em]$

Substituting value of BC from (1) in above equation :

$\Rightarrow \text{tan β} = \dfrac{BD}{\dfrac{x}{\text{tan α}}} \\[1em] \Rightarrow \text{tan β} = \dfrac{BA + AD}{\dfrac{x}{\text{tan α}}} \\[1em] \Rightarrow \text{tan β} \times \dfrac{x}{\text{tan α}} = x + h \\[1em] \Rightarrow x \text{tan β} = \text{tan α}(x + h) \\[1em] \Rightarrow x \text{tan β} = x \text{tan α} + h \text{tan α} \\[1em] \Rightarrow x \text{tan β} - x \text{tan α} = h \text{tan α} \\[1em] \Rightarrow x(\text{tan β} - \text{tan α}) = h \text{tan α} \\[1em] \Rightarrow x = \dfrac{\text{h tan α}}{\text{tan β} - \text{tan α}}.$

Hence, proved that the height of the tower = $\dfrac{\text{h tan α}}{\text{tan β} - \text{tan α}}.$

With reference to the given figure, a man stands on the ground at point A, which is on the same horizontal plane as B, the foot of the vertical pole BC. The height of the pole is 10 m. The man's eye is 2 m above the ground. He observes the angle of elevation of C, the top of the pole, as x°, where tan x° = $\dfrac{2}{5}$. Calculate:

(i) the distance AB in metres;

(ii) angle of elevation of the top of the pole when he is standing 15 metres from the pole. Give your answer to the nearest degree.

Answer

Let's take AD to be the height of the man, AD = 2 m.

From figure, BE = AD = 2 m.

Also,

CE = BC - BE = (10 - 2) = 8 m.

(i) In ∆CED,

$\Rightarrow \text{tan x} \degree = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan x} \degree = \dfrac{CE}{DE} \\[1em] \Rightarrow \dfrac{2}{5} = \dfrac{8}{DE} \\[1em] \Rightarrow DE = \dfrac{8 \times 5}{2} \\[1em] \Rightarrow DE = \dfrac{40}{2} = 20 \text{ m}.$

From figure,

AB = DE = 20 m.

Hence, AB = 20 m.

(ii) Let A'D' be the new position of the man and θ be the angle of elevation of the top of the tower.

Such that, D'E = 15 m

In ∆CED',

$\text{tan θ} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan θ} = \dfrac{CE}{D'E} \\[1em] \Rightarrow \text{tan θ} = \dfrac{8}{15} \\[1em] \Rightarrow \text{tan θ} = 0.533 \\[1em] \Rightarrow \text{tan θ} = \text{tan } 28° \\[1em] \Rightarrow \text{θ} = 28°.$

Hence, angle of elevation of the top of the pole when the man is standing 15 metres from the pole is 28°.

From a window A, 10 m above the ground the angle of elevation of the top C of a tower is x°, where tan x° = $\dfrac{5}{2}$ and the angle of depression of the foot D of the tower is y°, where tan y° = $\dfrac{1}{4}$. Calculate the height CD of the tower in metres.

Answer

From figure,

⇒ AB = DE = 10 m.

In ∆AED

$\Rightarrow \text{tan y°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan y°} = \dfrac{DE}{AE} \\[1em] \Rightarrow \dfrac{1}{4} = \dfrac{DE}{AE} \\[1em] \Rightarrow AE = 4DE = 4 \times 10 = 40 \text{ m}.$

In ∆AEC,

$\Rightarrow \text{tan x°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan x°} = \dfrac{CE}{AE} \\[1em] \Rightarrow \dfrac{5}{2} = \dfrac{CE}{AE} \\[1em] \Rightarrow CE = AE \times \dfrac{5}{2} = 40 \times \dfrac{5}{2} = 100 \text{ m}.$

From figure,

CD = DE + CE = 10 + 100 = 110 m.

Hence, height of tower (CD) = 110 m.

A vertical tower is 20 m high. A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is he standing from the foot of the tower ?

Answer

Let angle of elevation be θ.

According to question,

⇒ cos θ = 0.53

⇒ cos θ = cos 58°

⇒ θ = 58°.

⇒ cos² θ = 0.2809

⇒ 1 - sin² θ = 0.2809

⇒ sin² θ = 1 - 0.2809

⇒ sin² θ = 0.7191

⇒ sin θ = $\sqrt{0.7191}$

⇒ sin θ = 0.848

⇒ tan θ = $\dfrac{\text{sin θ}}{\text{cos θ}} = \dfrac{0.848}{0.53}$ = 1.6

$\therefore \dfrac{\text{Perpendicular}}{\text{Base}} = 1.6 \\[1em] \Rightarrow \dfrac{AB}{BC} = 1.6 \\[1em] \Rightarrow \dfrac{20}{BC} = 1.6 \\[1em] \Rightarrow BC = \dfrac{20}{1.6} = 12.5 \text{ m}.$

The man is standing at a distance of 12.5 meters.

A vertical pole and a vertical tower are on the same level ground in such a way that from the top of the pole the angle of elevation of the top of the tower is 60° and the angle of depression of the bottom of the tower is 30°. Find :

(i) the height of the tower, if the height of the pole is 20 m;

(ii) the height of the pole, if the height of the tower is 75 m

Answer

(i) Let AB be the pole and CD be the tower.

Given,

Length of pole (AB) = 20 m.

From figure,

CE = AB = 20 m.

In △AEC,

$\text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{CE}{AE} \\[1em] \Rightarrow AE = \sqrt{3} CE \\[1em] \Rightarrow AE = 20\sqrt{3} \text{ m}.$

In △AED,

$\text{tan 60°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \dfrac{DE}{AE} \\[1em] \Rightarrow DE = AE\sqrt{3} \\[1em] \Rightarrow DE = 20\sqrt{3} \times \sqrt{3} = 20 \times 3 = 60 \text{ m}.$

From figure,

CD = DE + CE = 60 + 20 = 80 m.

Hence, the height of the tower = 80 m.

(ii) Given,

Length of tower (CD) = 75 m

From figure,

AE = BC = y (let)

In △AEC,

$\text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{CE}{AE} \\[1em] \Rightarrow CE = \dfrac{AE}{\sqrt{3}} \\[1em] \Rightarrow CE = \dfrac{y}{\sqrt{3}} \text{ m}.$

In △AED,

$\text{tan 60°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \dfrac{DE}{AE} \\[1em] \Rightarrow DE = AE\sqrt{3} \\[1em] \Rightarrow DE = y\sqrt{3} \text{ m}.$

From figure,

⇒ CD = CE + DE

$\Rightarrow \dfrac{y}{\sqrt{3}} + \sqrt{3}y = 75 \\[1em] \Rightarrow \dfrac{y + 3y}{\sqrt{3}} = 75 \\[1em] \Rightarrow \dfrac{4y}{\sqrt{3}} = 75 \\[1em] \Rightarrow y = \dfrac{75\sqrt{3}}{4} \text{ m}.$

We know that,

CE = $\dfrac{y}{\sqrt{3}} = \dfrac{\dfrac{75\sqrt{3}}{4}}{\sqrt{3}} = \dfrac{75}{4}$ = 18.75 meters.

Hence, height of pole = 18.75 meters.

In the given figure, from the top of a building AB = 60 m high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30° and 60° respectively. Find :

(i) the horizontal distance between AB and CD.

(ii) the height of the lamp post.

Answer

(i) We know that,

Alternate angles are equal.

∴ ∠ACB = ∠EAC = 60°

In △ABC,

$\text{tan 60°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \dfrac{AB}{BC} \\[1em] \Rightarrow BC = \dfrac{AB}{\sqrt{3}} = \dfrac{60}{1.732} = 34.64 \text{ m}.$

Hence, horizontal distance between AB and CD = 34.64 meters.

(ii) We know that,

From figure,

FD = BC = 34.64 m

As, alternate angles are equal.

∴ ∠ADF = ∠EAD = 30°

In △AFD,

$\text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AF}{FD} \\[1em] \Rightarrow AF = \dfrac{FD}{\sqrt{3}} \\[1em] \Rightarrow AF = \dfrac{34.64}{1.732} = 20 \text{ m}.$

From figure,

BF = AB - AF = 60 - 20 = 40 m.

∴ CD = 40 m.

Hence, height of lamp post = 40 m.

An aeroplane, at an altitude of 250 m, observes the angle of depression of two boats on the opposite banks of a river to be 45° and 60° respectively. If the boats are on the opposite sides of the aeroplane, find the width of the river. Write the answer correct to the nearest whole number.

Answer

Let D be the position of aeroplane and A and B be the positions of boat.

As, angle of depression are 60° and 45°.

Since, alternate angles are equal.

∴ ∠DAC = EDA = 60° and ∠DBC = FDB = 45°.

In △ACD,

$\text{tan 60°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \dfrac{CD}{AC} \\[1em] \Rightarrow AC = \dfrac{CD}{\sqrt{3}} \\[1em] \Rightarrow AC = \dfrac{250}{1.732} \\[1em] \Rightarrow AC = 144.34 \text{ m}.$

In △BCD,

$\text{tan 45°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1 = \dfrac{CD}{BC} \\[1em] \Rightarrow BC = CD = 250 \text{ m}.$

AB = AC + BC = 144.34 + 250 = 394.34 ≈ 394 m.

Hence, width of river = 394 m.

The horizontal distance between two towers is 120 m. The angle of elevation of the top and angle of depression of the bottom of the first tower as observed from the top of the second tower is 30° and 24°, respectively. Find the height of the two towers. Give your answer correct to 3 significant figures.

Answer

From figure,

AB is the first tower and CD is the second tower.

From figure,

EC = BD = 120 m.

In △AEC,

$\text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AE}{EC} \\[1em] \Rightarrow AE = \dfrac{EC}{\sqrt{3}} \\[1em] \Rightarrow AE = \dfrac{120}{\sqrt{3}} \\[1em] \Rightarrow AE = \dfrac{120}{1.732} = 69.28 \text{ m}.$

In △EBC,

$\text{tan 24°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 0.445 = \dfrac{EB}{EC} \\[1em] \Rightarrow EB = EC \times 0.445 \\[1em] \Rightarrow EB = 120 \times 0.445 = 53.4 \text{ m}.$

From figure,

⇒ CD = EB = 53.4 meters.

⇒ AB = AE + EB = 69.28 + 53.4 = 122.68 ≈ 123 meters.

Hence, height of two towers = 123 meters and 53.4 meters.

The angles of depression of two ships A and B as observed from the top of a lighthouse 60 m high are 60° and 45° respectively. If the two ships are on the opposite sides of the lighthouse, find the distance between the two ships. Give your answer correct to nearest whole number.

Answer

Let CD be the lighthouse.

As, angle of depression are 60° and 45°.

Since, alternate angles are equal.

∴ ∠DAC = ∠EDA = 60° and ∠DBC = ∠FDB = 45°.

In △ACD,

$\text{tan 60°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \dfrac{CD}{AC} \\[1em] \Rightarrow AC = \dfrac{CD}{\sqrt{3}} \\[1em] \Rightarrow AC = \dfrac{60}{1.732} \\[1em] \Rightarrow AC = 34.64 \text{ m}.$

In △BCD,

$\text{tan 45°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1 = \dfrac{CD}{BC} \\[1em] \Rightarrow BC = CD = 60 \text{ m}.$

AB = AC + BC = 34.64 + 60 = 94.64 ≈ 95 m.

Hence, the distance between two ships = 95 m.

Two ships are approaching a light house from the opposite directions. The angles of depression of the two ships from the top of the light house are 30° and 45°. If the distance between the two ships is 100 metre, find the height of the light house correct to one decimal place.

Answer

Let height of lighthouse (CD) be h meters.

Let A and B be the boats approaching lighthouse.

From figure,

∠CAD = ∠ACX = 30° and ∠CBD = ∠BCY = 45°

In triangle ACD,

$\Rightarrow \tan 30^\circ = \dfrac{h}{x} \\[1em] \Rightarrow \dfrac{1}{\sqrt3} = \dfrac{h}{x} \\[1em] \Rightarrow x = h\sqrt3.$

In triangle BCD,

$\Rightarrow \tan 45^\circ = \dfrac{h}{y} \\[1em] \Rightarrow 1 = \dfrac{h}{y} \\[1em] \Rightarrow h = y.$

Given,

Distance between the two boats is 100 m.

x + y = 100

$\Rightarrow h\sqrt3 + h = 100 \\[1em] \Rightarrow h(\sqrt{3} + 1) = 100 \\[1em] \Rightarrow h =\dfrac{100}{\sqrt3 + 1} \\[1em] \Rightarrow h = \dfrac{100(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} \\[1em] \Rightarrow h = \dfrac{100(\sqrt{3} - 1)}{3 - 1} \\[1em] \Rightarrow h = \dfrac{100(\sqrt{3} - 1)}{2} \\[1em] \Rightarrow h = 50(\sqrt{3} - 1) \\[1em] \Rightarrow h = 50(1.732 - 1) \\[1em] \Rightarrow h = 50(0.732) \\[1em] \Rightarrow h = 36.6 \text{ m}.$

Hence, option 1 is the correct option.