Heights and Distances

The measure of x is :

1. $15\sqrt{3}$ cm

2. 15 cm

3. $5\sqrt{3}$ cm

4. 5 cm

Answer

We know that,

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

From figure,

In △ ABC,

⇒ tan 60° = $\dfrac{AB}{BC}$

⇒ $\sqrt{3} = \dfrac{15}{x}$

⇒ x = $\dfrac{15}{\sqrt{3}}$

Rationalizing the denominator,

$\Rightarrow x = \dfrac{15}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow x = \dfrac{15\sqrt{3}}{3} \\[1em] \Rightarrow x = 5\sqrt{3}.$

Hence, Option 3 is the correct option.

The value of x is :

1. 40 m

2. 30 m

3. 20 m

4. $40\sqrt{3}$ m

Answer

We know that,

cos θ = $\dfrac{\text{Base}}{\text{Hypotenuse}}$

From figure,

In △ ABC,

⇒ cos 60° = $\dfrac{BC}{AB}$

⇒ $\dfrac{1}{2} = \dfrac{x}{40}$

⇒ x = $\dfrac{40}{2}$ = 20 m.

Hence, Option 3 is the correct option.

In the given figure, AB = BD, BC = 20 cm and ∠D = 45°, the length of AC is :

1. 54.64 cm

2. 48.28 cm

3. 40 cm

4. 14.64 cm

Answer

We know that,

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

From figure,

In △ BCD,

⇒ sin 45° = $\dfrac{BC}{BD}$

⇒ $\dfrac{1}{\sqrt{2}} = \dfrac{20}{BD}$

⇒ BD = $20\sqrt{2}$ m.

From figure,

⇒ AB = BD = $20\sqrt{2}$ m

⇒ AC = AB + BC = $20\sqrt{2} + 20$ = 28.28 + 20 = 48.28 cm.

Hence, Option 2 is the correct option.

The measure of x correct to the nearest metre is :

1. 70 m

2. 35 m

3. 61 m

4. 140 m

Answer

We know that,

cos θ = $\dfrac{\text{Base}}{\text{Hypotenuse}}$

From figure,

In △ ABC,

⇒ cos 30° = $\dfrac{BC}{AB}$

⇒ $\dfrac{\sqrt{3}}{2} = \dfrac{x}{70}$

⇒ x = $\dfrac{70\sqrt{3}}{2} = 35\sqrt{3} = 60.62 \approx 61$ m.

Hence, Option 3 is the correct option.

The length of DC is :

1. 10 m

2. $10(\sqrt{3} + 1)$ m

3. 20 m

4. $10(\sqrt{3} - 1)$ m

Answer

We know that,

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

From figure,

In △ ABC,

⇒ tan 45° = $\dfrac{AB}{BC}$

⇒ $1 = \dfrac{10}{BC}$

⇒ BC = 10 m.

In △ ABD,

⇒ tan 30° = $\dfrac{AB}{BD}$

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

⇒ BD = $10\sqrt{3}$ m.

From figure,

DC = BD - BC = $10\sqrt{3} - 10 = 10(\sqrt{3} - 1)$ m.

Hence, Option 4 is the correct option.

The height of a tree is $\sqrt{3}$ times the length of its shadow. Find the angle of elevation of the sun.

Answer

Let AB be the tree and BC be the shadow of tree.

Let the length of the shadow (BC) of the tree be x meters.

So, the height of the tree (AB) = $\sqrt{3}x$ meters.

If θ is the angle of elevation of the sun, then we have :

In △ABC,

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

Hence, angle of elevation of sun is 60°.

The angle of elevation of the top of a tower from a point on the ground and at a distance of 160 m from its foot, is found to be 60°. Find the height of the tower.

Answer

Let AB be the tower and C be the point at a distance of 160 m from foot of tower.

Let,

AB = h meters.

From figure,

In △ABC,

$\Rightarrow \text{tan θ} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan 60°} = \dfrac{AB}{BC} \\[1em] \Rightarrow \sqrt{3} = \dfrac{h}{160} \\[1em] \Rightarrow h = 160\sqrt{3} \\[1em] \Rightarrow h = 160 \times 1.732 = 277.12 \text{ m}.$

Hence, the height of the tower is 277.12 m.

A ladder is placed along a wall such that its upper end is resting against a vertical wall. The foot of the ladder is 2.4 m from the wall and the ladder is making an angle of 68° with the ground. Find the height, up to which the ladder reaches.

Answer

Let AC be the ladder and height of wall upto which ladder reaches be h meters.

∴ AB = h meters and BC = 2.4 meters.

Given,

Ladder is making an angle of 68° with the ground.

From figure,

In △ABC,

$\Rightarrow \text{tan θ} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan 68°} = \dfrac{AB}{BC} \\[1em] \Rightarrow 2.475 = \dfrac{h}{2.4} \\[1em] \Rightarrow h = 2.4 \times 2.475 \\[1em] \Rightarrow h = 5.94 \text{ m}.$

Hence, the ladder reaches upto a height of 5.94 m.

Two persons are standing on the opposite sides of a tower. They observe the angles of elevation of the top of the tower to be 30° and 38° respectively. Find the distance between them, if the height of the tower is 50 m.

Answer

Let PQ be the tower.

Let one of the persons, A be at a distance of x meters and the second person B be at a distance of y metres from the foot of the tower (Q).

Given, that angle of elevation of A is 30°.

From figure,

In △PQA,

$\Rightarrow \text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{PQ}{AQ} \\[1em] \Rightarrow AQ = PQ\sqrt{3} = 50\sqrt{3} \\[1em] \Rightarrow x = 50 \times 1.732 \\[1em] \Rightarrow x = 86.6 \text{ meters}.$

Given, that angle of elevation of B is 38°.

In △PBQ,

$\Rightarrow \text{tan 38°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 0.7813 = \dfrac{PQ}{BQ} \\[1em] \Rightarrow BQ = \dfrac{50}{0.7813} \\[1em] \Rightarrow y = 64 \text{ meters}.$

AB = x + y = 86.6 + 64 = 150.6 meters.

Hence, the distance between two persons = 150.6 meters.

A boy, 1.6 m tall, is 20 m away from a tower and observes the angle of elevation of the top of the tower to be (i) 45° (ii) 60°. Find the height of the tower in each case.

Answer

(i) From figure,

When angle of elevation is 45°, then CE is the tower.

In △ADE,

$\Rightarrow \text{tan 45°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1 = \dfrac{DE}{AD} \\[1em] \Rightarrow AD = DE \\[1em] \Rightarrow DE = 20 \text{ meters}.$

From figure,

CD = AB.

CE = CD + DE = 1.6 + 20 = 21.6 meters.

Hence, height of tower when angle of elevation is 45° is 21.6 meters.

(ii) From figure,

When angle of elevation is 60°.

In △ADF,

$\Rightarrow \text{tan 60°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \dfrac{DF}{AD} \\[1em] \Rightarrow DF = AD\sqrt{3} \\[1em] \Rightarrow DF = 20 \times 1.732 = 34.64 \text{ meters}.$

From figure,

CD = AB.

CF = CD + DF = 1.6 + 34.64 = 36.24 meters.

Hence, height of tower when angle of elevation is 60° is 36.24 meters.

The upper part of a tree, broken over by the wind, makes an angle of 45° with the ground; and the distance from the root to the point where the top of the tree touches the ground, is 15 m. What was the height of the tree before it was broken?

Answer

Let A be the point from where tree breaks and C be the point where above part of tree touches the ground.

From figure,

In △ABC,

$\text{tan 45°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1 = \dfrac{AB}{BC} \\[1em] \Rightarrow AB = BC = 15 \text{ meters}.$

In right angle triangle ABC,

⇒ AC² = AB² + BC²

⇒ AC² = 15² + 15²

⇒ AC² = 225 + 225

⇒ AC² = 450

⇒ AC = $\sqrt{450} = 15\sqrt{2}$ meters.

Height of tree = AB + AC = 15 + $15\sqrt{2}$

= 15 + 21.21

= 36.21 meters.

Hence, height of tree before it was broken = 36.21 meters.

The angle of elevation of the top of an unfinished tower from a point at a distance of 80 m from its base is 30°. How much higher must the tower be raised so that its angle of elevation at the same point may be 60° ?

Answer

Let AB be the unfinished tower and C be the point 80 m from base of tower.

From figure,

In △ABC,

$\Rightarrow \text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{BC} \\[1em] \Rightarrow AB = \dfrac{BC}{\sqrt{3}} \\[1em] \Rightarrow AB = \dfrac{80}{\sqrt{3}} \\[1em] \Rightarrow AB = \dfrac{80}{1.732} \\[1em] \Rightarrow AB = 46.19\text{ meters}.$

Let tower be raised to point D in order to make angle of elevation 60°.

From figure,

In △DBC,

$\Rightarrow \text{tan 60°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \dfrac{BD}{BC} \\[1em] \Rightarrow BD = BC\sqrt{3} \\[1em] \Rightarrow BD = 80\sqrt{3} \\[1em] \Rightarrow BD = 138.56\text{ metres}.$

AD = BD - AB = 138.56 - 46.19 = 92.37 meters.

Hence, the tower must be raised by 92.37 meters.

At a particular time, when the sun's altitude is 30°, the length of the shadow of a vertical tower is 45 m. Calculate :

(i) the height of the tower,

(ii) the length of the shadow of the same tower, when the sun's altitude is :

(a) 45° (b) 60°.

Answer

(i) Let AB be the tower.

From figure,

In △ABC,

$\text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{BC} \\[1em] \Rightarrow \dfrac{h}{45} = \dfrac{1}{\sqrt{3}} \\[1em] \Rightarrow h = \dfrac{45}{\sqrt{3}} \\[1em] \Rightarrow h = \dfrac{45}{1.732} \\[1em] \Rightarrow h = 25.98 \text{ meters}.$

Hence, the height of tower = 25.98 meters.

(ii) When sun's altitude is

(a) 45°

$\Rightarrow \text{tan 45°} = \dfrac{AB}{BC} \\[1em] \Rightarrow 1 = \dfrac{h}{BC} \\[1em] \Rightarrow BC = h = 25.98 \text{ meters}.$

Hence, length of shadow of tower when the sun's altitude is 45° = 25.98 meters.

(b) 60°

$\Rightarrow \text{tan 60°} = \dfrac{AB}{BC} \\[1em] \Rightarrow \sqrt{3} = \dfrac{h}{BC} \\[1em] \Rightarrow BC = \dfrac{h}{\sqrt{3}} \\[1em] \Rightarrow BC = \dfrac{25.98}{1.732} \\[1em] \Rightarrow BC = 15\text{ meters}.$

Hence, length of shadow of tower when the sun's altitude is 60° = 15 meters.

Two vertical poles are on either side of a road. A 30 m long ladder is placed between the two poles. When the ladder rests against one pole, it makes angle 32° 24' with the pole and when it is turned to rest against another pole, it makes angle 32° 24' with the road. Calculate the width of the road.

Answer

Let AP and CQ be two poles.

When ladder is at position AB resting on pole AP.

Then, ∠BAP = 32° 24'

From figure,

In △ABP,

$\text{sin 32° 24}' = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] \Rightarrow 0.536 = \dfrac{BP}{AB} \\[1em] \Rightarrow BP = AB \times 0.536 \\[1em] \Rightarrow BP = 30 \times 0.536 \\[1em] \Rightarrow BP = 16.08 \text{ meters}.$

When ladder is at position BC resting on pole CQ.

Then it makes angle 32° 24' with road.

∴ ∠CBQ = 32° 24'

From figure,

In △BQC,

$\text{cos 32° 24}' = \dfrac{\text{Base}}{\text{Hypotenuse}} \\[1em] \Rightarrow 0.844 = \dfrac{BQ}{BC} \\[1em] \Rightarrow BQ = BC \times 0.844 \\[1em] \Rightarrow BQ = 30 \times 0.844 \\[1em] \Rightarrow BQ = 25.32\text{ meters}.$

Width of road = BP + BQ = 16.08 + 25.32 = 41.4 meters.

Hence, width of road = 41.4 meters.

Two climbers are at points A and B on a vertical cliff face. To an observer C, 40 m from the foot of the cliff, on the level ground, A is at an elevation of 48° and B of 57°. What is the distance between the climbers?

Answer

Let P be the foot of cliff.

From figure,

In △BCP,

$\text{tan 57°} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] \Rightarrow 1.539 = \dfrac{BP}{PC} \\[1em] \Rightarrow BP = PC \times 1.539 \\[1em] \Rightarrow BP = 40 \times 1.539 \\[1em] \Rightarrow BP = 61.57 \text{ meters}.$

In △ACP,

$\text{tan 48°} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] \Rightarrow 1.110 = \dfrac{AP}{PC} \\[1em] \Rightarrow AP = PC \times 1.110 \\[1em] \Rightarrow AP = 40 \times 1.110 \\[1em] \Rightarrow AP = 44.40 \text{ meters}.$

Distance between climbers (BA) = BP - AP

= 61.57 - 44.40

= 17.17 meters.

Hence, distance between two climbers = 17.17 meters.

A man stands 9 m away from a flag-pole. He observes that angle of elevation of the top of the pole is 28° and the angle of depression of the bottom of the pole is 13°. Calculate the height of pole.

Answer

Let AC be the pole and D be the point where man stands.

From figure,

In △ADB,

$\Rightarrow \text{tan 28°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 0.532 = \dfrac{AB}{BD} \\[1em] \Rightarrow AB = BD \times 0.532 \\[1em] \Rightarrow AB = 9 \times 0.532 = 4.788 \text{ m}.$

In △BDC,

$\Rightarrow \text{tan 13°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 0.231 = \dfrac{BC}{BD} \\[1em] \Rightarrow BC = BD \times 0.231 \\[1em] \Rightarrow BC = 9 \times 0.231 = 2.079 \text{ m}.$

AC = AB + BC = 4.788 + 2.079 = 6.867 meters.

Hence, the height of pole = 6.867 meters.

From the top of a cliff 92 m high, the angle of depression of a buoy is 20°. Calculate, to the nearest metre, the distance of the buoy from the foot of the cliff.

Answer

Let AB be the cliff and C be the buoy.

Given,

AB = 92 m

From figure,

In △ACB,

$\text{tan 20°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan 20°} = \dfrac{AB}{BC} \\[1em] \Rightarrow 0.364 = \dfrac{92}{BC}\\[1em] \Rightarrow BC = \dfrac{92}{0.364} \\[1em] \Rightarrow BC = 252.7 ≈ 253 \text{ meters}.$

Hence, the distance of the buoy from the foot of the cliff is 253 meters.

According to the information given in the following figure, the length of BC is :

1. $\sqrt{3}$ m

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

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

4. $2\sqrt{3}$ m

Answer

We know that,

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

From figure,

In △ ABD,

⇒ tan 45° = $\dfrac{AD}{BD}$

⇒ $1 = \dfrac{1}{BD}$

⇒ BD = 1 m.

In △ ACD,

⇒ tan 30° = $\dfrac{AD}{CD}$

⇒ $\dfrac{1}{\sqrt{3}} = \dfrac{1}{CD}$

⇒ CD = $\sqrt{3}$ m.

From figure,

BC = BD + CD = $(1 + \sqrt{3})$ m.

Hence, Option 3 is the correct option.

The measurement of h is :

1. $\sqrt{8 \times 6}$ cm

2. $\sqrt{\dfrac{4}{3}}$ cm

3. (8 - 6) cm

4. $2\sqrt{2}$ cm

Answer

We know that,

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

From figure,

In △ ABC,

⇒ tan 50° = $\dfrac{AB}{BC}$

⇒ tan 50° = $\dfrac{h}{6}$

⇒ h = 6 tan 50°

⇒ h = 6 tan (90° - 40°)

⇒ h = 6 cot 40° ........(1)

In △ ABD,

⇒ tan 40° = $\dfrac{AB}{BD}$

⇒ tan 40° = $\dfrac{h}{8}$

⇒ h = 8 tan 40° ........(2)

Multiplying equation (1) and (2), we get :

⇒ h × h = 6 cot 40° × 8 tan 40°

⇒ h² = 8 × 6 × cot 40° $\times \dfrac{1}{\text{cot 40°}}$

⇒ h² = 8 × 6

⇒ h = $\sqrt{8 \times 6}$.

Hence, Option 1 is the correct option.

The length of AC is :

1. 22 m

2. 38 m

3. 23 m

4. 45 m

Answer

From figure,

BCDE is a rectangle.

∴ BC = DE = 8 m and BE = DC = 30 m.

We know that,

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

From figure,

In △ ABE,

⇒ tan 45° = $\dfrac{AB}{BE}$

⇒ $1 = \dfrac{AB}{30}$

⇒ AB = 30 m.

From figure,

AC = AB + BC = 30 + 8 = 38 m.

Hence, Option 2 is the correct option.

Using the information given in the following figure, the measurement of AE is :

1. 20 cm

2. 10 cm

3. $6\dfrac{2}{3}$ cm

4. $20\sqrt{3}$ cm

Answer

We know that,

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

From figure,

In △ ABC,

⇒ tan 60° = $\dfrac{AB}{BC}$

⇒ $\sqrt{3} = \dfrac{20}{BC}$

⇒ BC = $\dfrac{20}{\sqrt{3}}$ cm.

In rectangle BCDE,

Opposite sides of rectangle are equal.

∴ DE = BC = $\dfrac{20}{\sqrt{3}}$ cm

From figure,

In △ AED,

⇒ tan 30° = $\dfrac{AE}{DE}$

⇒ $\dfrac{1}{\sqrt{3}} = \dfrac{AE}{\dfrac{20}{\sqrt{3}}}$

⇒ AE = $\dfrac{1}{\sqrt{3}} \times \dfrac{20}{\sqrt{3}} = \dfrac{20}{3} = 6\dfrac{2}{3}$ cm.

Hence, Option 3 is the correct option.

The length of DC is :

1. $15\sqrt{3}$ cm

2. $15(\sqrt{3} - 1)$ cm

3. $15(\sqrt{3} + 1)$ cm

4. $10\sqrt{3}$ cm

Answer

We know that,

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

From figure,

In △ ABC,

⇒ tan 60° = $\dfrac{AB}{BC}$

⇒ $\sqrt{3} = \dfrac{15}{BC}$

⇒ BC = $\dfrac{15}{\sqrt{3}}$

Rationalizing the denominator,

$\Rightarrow \text{BC } = \dfrac{15}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow \text{BC } = \dfrac{15\sqrt{3}}{3} \\[1em] \Rightarrow \text{BC } = 5\sqrt{3}.$

In △ ABD,

⇒ tan 30° = $\dfrac{AB}{BD}$

⇒ $\dfrac{1}{\sqrt{3}} = \dfrac{15}{BD}$

⇒ BD = $15\sqrt{3}$ cm.

From figure,

DC = BD - BC = $15\sqrt{3} - 5\sqrt{3} = 10\sqrt{3}$ cm.

Hence, Option 4 is the correct option.

In the figure, given below, it is given that AB is perpendicular to BD and is of length X metres. DC = 30 m, ∠ADB = 30° and ∠ACB = 45°. Find X.

Answer

From figure,

In △ABD,

$\Rightarrow \text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{BD} \\[1em] \Rightarrow AB = \dfrac{BD}{\sqrt{3}} ........(1)$

In △ABC,

$\Rightarrow \text{tan 45°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1 = \dfrac{AB}{BC} \\[1em] \Rightarrow AB = BC ..........(2)$

From (1) and (2),

$\Rightarrow BC = \dfrac{BD}{\sqrt{3}} \\[1em] \Rightarrow \sqrt{3}BC = BD \\[1em] \Rightarrow \sqrt{3}BC = BC + CD \\[1em] \Rightarrow \sqrt{3}BC - BC = 30 \\[1em] \Rightarrow BC(\sqrt{3} - 1) = 30 \\[1em] \Rightarrow BC = \dfrac{30}{\sqrt{3} - 1} \\[1em] \Rightarrow BC = \dfrac{30}{1.732 - 1} \\[1em] \Rightarrow BC = \dfrac{30}{0.732} \\[1em] \Rightarrow BC = 40.98 \text{ meters}.$

∴ AB = X = 40.98 meters .....[From (2)]

Hence, X = 40.98 meters.

Find the height of a tree when it is found that on walking away from it 20 m, in a horizontal line through its base, the elevation of its top changes from 60° to 30°.

Answer

Let AB be the tree.

Let the two points be C and D such that CD = 20 m, ∠ADB = 30° and ∠ACB = 60°.

In ∆ABC,

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

In ∆ABD,

$\Rightarrow \text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{BD} \\[1em] \Rightarrow AB = \dfrac{BD}{\sqrt{3}} ..........(2)$

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

$\Rightarrow BC\sqrt{3} = \dfrac{BD}{\sqrt{3}} \\[1em] \Rightarrow BD = 3BC \\[1em] \Rightarrow BC + CD = 3BC \\[1em] \Rightarrow CD = 2BC \\[1em] \Rightarrow BC = \dfrac{CD}{2} \\[1em] \Rightarrow BC = \dfrac{20}{2} \\[1em] \Rightarrow BC = 10 \text{ meters}.$

AB = $BC \times \sqrt{3}$ = BC × 1.732 = 17.32 meters.

Hence, the height of the tree is 17.32 metres.

From the top of a light house 100 m high, the angles of depression of two ships are observed as 48° and 36° respectively. Find the distance between the two ships (in the nearest metre) if:

(i) the ships are on the same side of the light house.

(ii) the ships are on the opposite sides of the light house.

Answer

(i) Let's consider AB to be the lighthouse.

Given, depression angles are 48° and 36°.

When ships are on the same side,

In ∆ABC,

$\text{tan 48°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1.1106 = \dfrac{AB}{BC} \\[1em] \Rightarrow 1.1106 = \dfrac{100}{BC} \\[1em] \Rightarrow BC = \dfrac{100}{1.1106} \\[1em] \Rightarrow BC = 90.04 \text{ meters}.$

In ∆ABD,

$\text{tan 36°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 0.7265 = \dfrac{AB}{BD} \\[1em] \Rightarrow 0.7265 = \dfrac{100}{BD} \\[1em] \Rightarrow BD = \dfrac{100}{0.7265} \\[1em] \Rightarrow BD = 137.64 \text{ meters}.$

Distance between the two ships (CD) = BD – BC = 137.64 - 90.04

= 47.6 ≈ 48 m.

Hence, distance between ships when on the same side = 48 m.

(ii) Let's consider AB to be the lighthouse.

Given, depression angles are 48° and 36°.

As, alternate angles are equal.

∴ ∠ADB = ∠QAD = 36° and ∠ACB = ∠PAC = 48°.

When ships are on the opposite side,

In ∆ABC,

$\text{tan 48°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1.1106 = \dfrac{AB}{BC} \\[1em] \Rightarrow 1.1106 = \dfrac{100}{BC} \\[1em] \Rightarrow BC = \dfrac{100}{1.1106} \\[1em] \Rightarrow BC = 90.04 \text{ meters}.$

In ∆ABD,

$\text{tan 36°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 0.7265 = \dfrac{AB}{BD} \\[1em] \Rightarrow 0.7265 = \dfrac{100}{BD} \\[1em] \Rightarrow BD = \dfrac{100}{0.7265} \\[1em] \Rightarrow BD = 137.64 \text{ meters}.$

Distance between the two ships (CD) = BD + BC = 137.64 + 90.04

= 227.68 ≈ 228 m.

Hence, the distance between two ships, when on opposite side = 228 m.

Two pillars of equal heights stand on either side of a roadway, which is 150 m wide. At a point in the roadway between the pillars the elevations of the tops of the pillars are 60° and 30°; find the height of the pillars and the position of the point.

Answer

Let AB and CD be the two towers of height h meters. Let P be a point in the roadway BD such that BD = 150 m, ∠APB = 60° and ∠CPD = 30°.

In ∆ABP,

$\text{tan 60°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \dfrac{AB}{BP} \\[1em] \Rightarrow BP = \dfrac{AB}{\sqrt{3}} \\[1em] \Rightarrow BP = \dfrac{h}{\sqrt{3}}\space ............(1)$

In ∆CDP,

$\text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{CD}{PD} \\[1em] \Rightarrow PD = \sqrt{3}CD \\[1em] \Rightarrow PD = \sqrt{3}h\space ............(2)$

We know that,

⇒ BD = 150 m

⇒ BP + PD = 150 m

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

$\Rightarrow \dfrac{h}{\sqrt{3}} + \sqrt{3}h = 150 \\[1em] \Rightarrow \dfrac{h + 3h}{\sqrt{3}} = 150 \\[1em] \Rightarrow \dfrac{4h}{\sqrt{3}} = 150 \\[1em] \Rightarrow 4h = 150 \times \sqrt{3} \\[1em] \Rightarrow h = \dfrac{150 \times 1.732}{4} \\[1em] \Rightarrow h = 37.5 \times 1.732 \\[1em] \Rightarrow h = 64.95 \text{ meters}.$

From equation (1),

BP = $\dfrac{h}{\sqrt{3}} = \dfrac{64.95}{1.732}$ = 37.5 meters.

Hence, height of each pillar is 64.95 m and the point P is 37.5 m from the pillar AB.

From the figure, given below, calculate the length of CD.

Answer

From figure,

DE = CB = 15 m.

In ∆AED,

$\text{tan 22°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 0.404 = \dfrac{AE}{DE} \\[1em] \Rightarrow 0.404 = \dfrac{AE}{15} \\[1em] \Rightarrow AE = 15 \times 0.404 \\[1em] \Rightarrow AE = 6.06 \text{ meters}.$

In ∆ABC,

$\text{tan 47°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1.0724 = \dfrac{AB}{BC} \\[1em] \Rightarrow 1.0724 = \dfrac{AB}{15} \\[1em] \Rightarrow AB = 15 \times 1.0724 \\[1em] \Rightarrow AB = 16.086 \text{ meters}.$

From figure,

CD = BE = AB - AE

= 16.086 - 6.06

= 10.03 meters.

Hence, CD = 10.03 meters.

The angle of elevation of the top of a tower is observed to be 60°. At a point, 30 m vertically above the first point of observation, the elevation is found to be 45°. Find:

(i) the height of the tower,

(ii) its horizontal distance from the points of observation.

Answer

(i) Let AB be the the tower, C be the first point of observation and D be the second point.

From figure,

BC = ED = a (let) and BE = CD = 30 m.

In △ABC,

$\text{tan 60°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \dfrac{AB}{BC}\\[1em] \Rightarrow AB = \sqrt{3}BC \\[1em] \Rightarrow AB = \sqrt{3}a ..........(1)$

In △AED,

$\text{tan 45°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1 = \dfrac{AE}{ED}\\[1em] \Rightarrow AE = ED \\[1em] \Rightarrow AE = a ..........(2)$

We know that,

⇒ CD = AB - AE

⇒ 30 = $\sqrt{3}a - a$

⇒ 30 = $a(\sqrt{3} - 1)$

⇒ a = $\dfrac{30}{\sqrt{3} - 1} = \dfrac{30}{1.732 - 1} = \dfrac{30}{0.732}$ = 40.98 metres.

From equation (1),

AB = $\sqrt{3}a = 1.732 \times$ 40.98 = 70.98 meters.

Hence, height of tower = 70.98 meters.

(ii) From part (i),

ED = a = 40.98 meters.

Hence, horizontal distance from the points of observation is 40.98 meters.

From the top of a cliff, 60 meters high, the angles of depression of the top and bottom of a tower are observed to be 30° and 60°. Find the height of the tower.

Answer

Let CD be the cliff so CD = 60 meters and AB be the tower.

Since, alternate angles are equal,

∴ ∠ECA = ∠CAF = 30° and ∠ECB = ∠CBD = 60°.

Let AF = BD = a meters.

In △BCD,

$\text{tan 60°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \dfrac{CD}{BD}\\[1em] \Rightarrow CD = \sqrt{3}BD \\[1em] \Rightarrow CD = \sqrt{3}a \\[1em] \Rightarrow 60 = \sqrt{3}a \\[1em] \Rightarrow a = \dfrac{60}{\sqrt{3}} \\[1em] \Rightarrow a = \dfrac{60}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow a = \dfrac{60\sqrt{3}}{3} \\[1em] \Rightarrow a = 20\sqrt{3} \text{ meters}.$

In △AFC,

$\text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{CF}{AF}\\[1em] \Rightarrow CF = \dfrac{AF}{\sqrt{3}} \\[1em] \Rightarrow CF = \dfrac{a}{\sqrt{3}} \\[1em] \Rightarrow CF = \dfrac{20\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow CF = 20 \text{ meters}.$

From figure,

⇒ AB = DF = CD - CF

⇒ AB = CD - CF

⇒ AB = 60 - 20 = 40 meters.

Hence, the height of tower = 40 meters.

A man on a cliff observes a boat, at an angle of depression 30°, which is sailing towards the shore to the point immediately beneath him. Three minutes later, the angle of depression of the boat is found to be 60°. Assuming that the boat sails at a uniform speed, determine :

(i) how much more time it will take to reach the shore ?

(ii) the speed of the boat in metre per second, if the height of the cliff is 500 m.

Answer

Let CD be the cliff and A be the position of the ship when angle of elevation is 30° and B be the position when angle of elevation is 60°.

(i) From figure,

$\text{tan 60°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \dfrac{CD}{BD} \space .............(1) \\[1em] \text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{CD}{AD} \space .............(2)$

Dividing equation (1) by (2), we get :

$\Rightarrow \dfrac{\sqrt{3}}{\dfrac{1}{\sqrt{3}}} = \dfrac{\dfrac{CD}{BD}}{\dfrac{CD}{AD}} \\[1em] \Rightarrow \sqrt{3} \times \sqrt{3} = \dfrac{AD}{BD} \\[1em] \Rightarrow \dfrac{AD}{BD} = 3 \\[1em] \Rightarrow AD = 3BD \\[1em] \Rightarrow AB + BD = 3BD \\[1em] \Rightarrow AB = 2BD.$

Boat reaches from point A to B in 3 minutes.

Thus, boat covers distance AB in 3 minutes or it covers distance

⇒ 2 BD in 3 minutes

⇒ BD in $\dfrac{3}{2}$ = 1.5 minutes.

Hence, it will takes 1.5 minutes more to reach the shore.

(ii) In △ADC,

$\text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{CD}{AD} \\[1em] \Rightarrow AD = \sqrt{3}CD$

In △BDC,

$\text{tan 60°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \dfrac{CD}{BD} \\[1em] \Rightarrow BD = \dfrac{CD}{\sqrt{3}}$

From figure,

$\Rightarrow AB = AD - BD \\[1em] \Rightarrow AB = \sqrt{3}CD - \dfrac{CD}{\sqrt{3}} \\[1em] \Rightarrow AB = \dfrac{3CD - CD}{\sqrt{3}} \\[1em] \Rightarrow AB = \dfrac{2CD}{\sqrt{3}} .........(1)$

Let speed of boat be a meter/second so in 3 minutes boat will travel :

Distance (AB) = Speed × Time

AB = a × 3 × 60

AB = 180a meters .........(2)

From (1) and (2) we get,

$\Rightarrow \dfrac{2CD}{\sqrt{3}} = 180a \\[1em] \Rightarrow \dfrac{2 \times 500}{\sqrt{3}} = 180a \\[1em] \Rightarrow \dfrac{1000}{1.732} = 180a \\[1em] \Rightarrow a = \dfrac{1000}{180 \times 1.732} \\[1em] \Rightarrow a = \dfrac{1000}{311.76} = 3.21 \text{ m/s}.$

Hence, speed of boat = 3.21 m/s.

A man in a boat rowing away from a lighthouse 150 m high, takes 2 minutes to change the angle of elevation of the top of the lighthouse from 60° to 45°. Find the speed of the boat.

Answer

Let man in the boat be originally at point C and after 2 minutes it reaches the point D and AB be the lighthouse.

AB = 150 meters.

In △ABC,

$\text{tan 60°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \dfrac{AB}{BC} \\[1em] \Rightarrow AB = \sqrt{3} BC \\[1em] \Rightarrow BC = \dfrac{AB}{\sqrt{3}} \\[1em] \Rightarrow BC = \dfrac{150}{\sqrt{3}} \\[1em] \Rightarrow BC = \dfrac{150}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow BC = \dfrac{150\sqrt{3}}{3} \\[1em] \Rightarrow BC = 50\sqrt{3} \text{ meters}.$

In △ABD,

$\text{tan 45°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1 = \dfrac{AB}{BD} \\[1em] \Rightarrow AB = BD \\[1em] \Rightarrow BD = 150 \text{ metres}.$

CD = BD - BC = 150 - $50\sqrt{3}$

= 150 - 86.6

= 63.4 meters.

In 2 minutes boat covers 63.4 meters or boat covers 63.4 meters in 120 seconds.

Speed = $\dfrac{\text{Distance}}{\text{Time}} = \dfrac{63.4}{120}$ = 0.53 m/sec.

Hence, the speed of boat = 0.53 m/sec.

A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60°. When he moves 40 m away from the bank, he finds the angle of elevation to be 30°. Find :

(i) the height of the tree, correct to 2 decimal places.

(ii) the width of the river.

Answer

Let CD be the tree and B be the position of the person when angle of elevation is 60° and A be the position when angle of elevation is 30°.

(i) In △BCD,

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

In △ACD,

$\text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{CD}{AC} \\[1em] \Rightarrow AC = CD\sqrt{3}.$

From figure,

AB = AC - BC

$\Rightarrow 40 = CD\sqrt{3} - \dfrac{CD}{\sqrt{3}} \\[1em] \Rightarrow \dfrac{3CD - CD}{\sqrt{3}} = 40 \\[1em] \Rightarrow \dfrac{2CD}{\sqrt{3}} = 40 \\[1em] \Rightarrow CD = \dfrac{40\sqrt{3}}{2} = 20\sqrt{3} \\[1em] \Rightarrow CD = 20 \times 1.732 = 34.64 \text{ m}.$

Hence, height of tree = 34.64 meters.

(ii) From part (i), we get :

BC = $\dfrac{CD}{\sqrt{3}} = \dfrac{20\sqrt{3}}{\sqrt{3}}$

= 20 meters.

Hence, width of river = 20 meters.

The horizontal distance between two towers is 75 m and the angular depression of the top of the first tower as seen from the top of the second, which is 160 m high, is 45°. Find the height of the first tower.

Answer

Let AB be the first tower and CD be the second tower and ∠EDA = 45° is the angle of depression.

Given, angle of depression of the top of the first tower as seen from the top of the second tower is 45°.

We know that,

Alternate angles are equal.

∴ ∠DAF = ∠EDA = 45°.

From figure,

AF = BC = 75 m.

In △ADF,

$\text{tan 45°}= \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1 = \dfrac{DF}{AF} \\[1em] \Rightarrow DF = AF = 75 \text{ m}.$

From figure,

AB = FC = CD - DF = 160 - 75 = 85 m.

Hence, height of first tower = 85 m.

The length of the shadow of a tower standing on level plane is found to be 2y meters longer when the sun's altitude is 30° than when it was 45°. Prove that the height of the tower is $y(\sqrt{3} + 1)$ meters.

Answer

Let CD be the tower of height h meters and BC be shadow when angle of elevation is 45° and AC be the shadow when angle of elevation is 30°.

In △ACD,

$\text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{CD}{AC} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{h}{AC} \\[1em] \Rightarrow h = \dfrac{AC}{\sqrt{3}} \text{ m} .........(1)$

In △BCD,

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

AC = AB + BC = (2y + h) meters.

Substituting value of AC in equation 1, we get :

$\Rightarrow h = \dfrac{2y + h}{\sqrt{3}} \\[1em] \Rightarrow \sqrt{3}h = 2y + h \\[1em] \Rightarrow \sqrt{3}h - h = 2y \\[1em] \Rightarrow h(\sqrt{3} - 1) = 2y \\[1em] \Rightarrow h = \dfrac{2y}{\sqrt{3} - 1}.$

Multiplying numerator and denominator by $(\sqrt{3} + 1)$.

$\Rightarrow h = \dfrac{2y}{\sqrt{3} - 1} \times \dfrac{\sqrt{3} + 1}{\sqrt{3} + 1} \\[1em] \Rightarrow h = \dfrac{2y(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \\[1em] \Rightarrow h = \dfrac{2y(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2} \\[1em] \Rightarrow h = \dfrac{2y(\sqrt{3} + 1)}{3 - 1} \\[1em] \Rightarrow h = \dfrac{2y(\sqrt{3} + 1)}{2} \\[1em] \Rightarrow h = y(\sqrt{3} + 1).$

Hence, proved that the height of tower = $y(\sqrt{3} + 1)$ meters.

An aeroplane flying horizontally 1 km above the ground and going away from the observer is observed at an elevation of 60°. After 10 seconds, its elevation is observed to be 30°; find the uniform speed of the aeroplane in km per hour.

Answer

Let aeroplane be originally at point E and after 10 seconds it reaches point D.

In △ABE,

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

In △ACD,

$\text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{CD}{AC} \\[1em] \Rightarrow AC = \sqrt{3}CD \\[1em] \Rightarrow AC = \sqrt{3} \times 1 = \sqrt{3} \text{ km}.$

From figure,

DE = BC.

BC = AC - AB = $\sqrt{3} - \dfrac{1}{\sqrt{3}}$

= $\dfrac{3 - 1}{\sqrt{3}}$

= $\dfrac{2}{1.732}$

= 1.1547 km.

∴ DE = 1.1547 km.

∴ Aeroplane travels 1.1547 km in 10 seconds.

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

Speed = $\dfrac{\text{Distance}}{\text{Time}} = \dfrac{1.1547}{\dfrac{1}{360}}$

= 1.1547 × 360

= 415.67 km/hr.

Hence, speed of aeroplane = 415.67 km/hr.

From the top of a hill, the angles of depression of two consecutive kilometer stones, due east are found to be 30° and 45° respectively. Find the distances of the two stones from the foot of hill.

Answer

Let C and D be the position of two kilometer stones and AB be the hill.

In △ABD,

$\text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{BD} \\[1em] \Rightarrow BD = \sqrt{3} AB .........(1)$

In △ABC,

$\text{tan 45°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1 = \dfrac{AB}{BC} \\[1em] \Rightarrow BC = AB ............(2)$

From figure,

⇒ CD = BD - BC

⇒ 1 = $\sqrt{3}AB - AB$

⇒ 1 = $AB(\sqrt{3} - 1)$

⇒ AB = $\dfrac{1}{\sqrt{3} - 1} = \dfrac{1}{0.732}$ = 1.366 km.

From equation (2),

BC = AB = 1.366 km

BD = BC + CD = 1.366 + 1 = 2.366 km.

Hence, kilometer stones are at a distance of 1.366 and 2.366 km.

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.