Rs Aggarwal 2021 2022 for Class 10 Maths Chapter 14 - Heights And Distances

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Rs Aggarwal 2021 2022 Solutions for Class 10 Maths Chapter 14 Heights And Distances are provided here with simple step-by-step explanations. These solutions for Heights And Distances are extremely popular among Class 10 students for Maths Heights And Distances Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2021 2022 Book of Class 10 Maths Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2021 2022 Solutions. All Rs Aggarwal 2021 2022 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

Page No 656:

Answer:

Let $A B$ be the tower standing vertically on the ground and O be the position of the observer.
We now have:
$O A = 20 m,$ ∠ $O A B = 90^{o}$ and ∠ $A O B$ $= 60^{o}$
Let:
$A B = h$ m

Now, in the right ∆ $O A B$ , we have:
$\frac{A B}{O A} = \tan 60^{o}$ = $\sqrt{3}$

⇒ $\frac{h}{20} = \sqrt{3}$

⇒ $h = 20 \sqrt{3}$ = $(20 \times 1.732) = 34.64$

Hence, the height of the pole is 34.64 m.

Page No 657:

Answer:

Let $O X$ be the horizontal ground and $A$ be the position of the kite.
Also, let O be the position of the observer and $O A$ be the thread.
Now, draw $A B$ ⊥ $O X$ .
We have:
∠ $B O A =$ $60^{o}$ , $O A = 75$ m and ∠ $O B A = 90^{o}$
Height of the kite from the ground = $A B$ = 75 m
Length of the string, $O A = x$ m

In the right ∆ $O B A$ , we have:
$\frac{A B}{O A} = \sin 60^{o} = \frac{\sqrt{3}}{2}$
⇒ $\frac{75}{x} = \frac{\sqrt{3}}{2}$
⇒ $x = \frac{75 \times 2}{\sqrt{3}} = \frac{150}{1.732} = 86.6$ m
Hence, the length of the string is $86.6$ m.

Page No 657:

Answer:

Let CE and AD be the heights of the observer and the chimney, respectively.

We have,
$BD = CE = 1.5 m, BC = DE = 30 m and ∠ ACB = 60 ° In ∆ ABC, \tan 60 ° = \frac{AB}{BC} \Rightarrow \sqrt{3} = \frac{AD - BD}{30} \Rightarrow AD - 1.5 = 30 \sqrt{3} \Rightarrow AD = 30 \sqrt{3} + 1.5 \Rightarrow AD = 30 \times 1.732 + 1.5 \Rightarrow AD = 51.96 + 1.5 \Rightarrow AD = 53.46 m$

So, the height of the chimney is 53.46 m (approx.).

Page No 657:

Answer:

Let the height of the tower be AB.

We have,
$AC = 5 m, AD = 20 m Let the angle of elevation of the top of the tower (i . e . ∠ ACB) from point C be θ . Then, the angle of elevation of the top of the tower (i . e . ∠ ADB) from point D = (90 ° - θ) Now, in ∆ ABC, \tan θ = \frac{AB}{AC} \Rightarrow \tan θ = \frac{AB}{5} . . . . . (i) Also, in ∆ ABD, \cot (90 ° - θ) = \frac{AD}{AB} \Rightarrow \tan θ = \frac{20}{AB} . . . . . (ii) From (i) and (ii), we get \frac{AB}{5} = \frac{20}{AB} \Rightarrow {AB}^{2} = 100 \Rightarrow AB = \sqrt{100} ∴ AB = 10 m$

So, the height of the tower is 10 m.

Page No 657:

Answer:

Let BC and CD be the heights of the tower and the flagstaff, respectively.

We have,
$AB = 120 m, ∠ BAC = 45 °, ∠ BAD = 60 ° Let CD = x In ∆ ABC, \tan 45 ° = \frac{BC}{AB} \Rightarrow 1 = \frac{BC}{120} \Rightarrow BC = 120 m Now, in ∆ ABD, \tan 60 ° = \frac{BD}{AB} \Rightarrow \sqrt{3} = \frac{BC + CD}{120} \Rightarrow BC + CD = 120 \sqrt{3} \Rightarrow 120 + x = 120 \sqrt{3} \Rightarrow x = 120 \sqrt{3} - 120 \Rightarrow x = 120 (\sqrt{3} - 1) \Rightarrow x = 120 (1.732 - 1) \Rightarrow x = 120 (0.732) \Rightarrow x = 87.84 \approx 87.8 m$

So, the height of the flagstaff is 87.8 m.

Page No 657:

Answer:

Let BC be the tower and CD be the water tank.

$We have, AB = 40 m, ∠ BAC = 30 ° and ∠ BAD = 45 ° In ∆ ABD, \tan 45 ° = \frac{BD}{AB} \Rightarrow 1 = \frac{BD}{40} \Rightarrow BD = 40 m Now, in ∆ ABC, \tan 30 ° = \frac{BC}{AB} \Rightarrow \frac{1}{\sqrt{3}} = \frac{BC}{40} \Rightarrow BC = \frac{40}{\sqrt{3}} \Rightarrow BC = \frac{40}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \Rightarrow BC = \frac{40 \sqrt{3}}{3} m (i) The height of the tower, BC = \frac{40 \sqrt{3}}{3} = \frac{40 \times 1.73}{3} = 23.067 \approx 23.1 m (ii) The depth of the tank, CD = (BD - BC) = (40 - 23.1) = 16.9 m$

Page No 657:

Answer:

Let AB be the tower and BC be the flagstaff.

We have,
$BC = 6 m, ∠ AOB = 30 ° and ∠ AOC = 60 ° Let AB = h In ∆ AOB, \tan 30 ° = \frac{AB}{OA} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{OA} \Rightarrow OA = h \sqrt{3} . . . . . (i) Now, in ∆ AOC, \tan 60 ° = \frac{AC}{OA} \Rightarrow \sqrt{3} = \frac{AB + BC}{h \sqrt{3}} [Using (i)] \Rightarrow 3 h = h + 6 \Rightarrow 3 h - h = 6 \Rightarrow 2 h = 6 \Rightarrow h = \frac{6}{2} \Rightarrow h = 3 m$

So, the height of the tower is 3 m.

Page No 657:

Answer:

Let $A C$ be the pedestal and $B C$ be the statue such that $B C = 1.46$ m.
We have:
∠ $A D C = 45^{o}$ and ∠ $A D B = 60^{o}$
Let:
$A C = h$ m and $A D = x$ m

In the right ∆ $A D C$ , we have:
$\frac{A C}{A D} = \tan 45^{o} = 1$

⇒ $\frac{h}{x} = 1$
⇒ $h = x$
Or,
$x = h$

Now, in the right ∆ $A D B$ , we have:
$\frac{A B}{A D} = \tan 60^{o} = \sqrt{3}$

⇒ $\frac{h + 1.46}{x} = \sqrt{3}$

On putting $x = h$ in the above equation, we get:
$\frac{h + 1.46}{h} = \sqrt{3}$

⇒ $h + 1.46 = \sqrt{3} h$
⇒ $h (\sqrt{3} - 1) = 1.46$
⇒ $h = \frac{1.46}{(\sqrt{3} - 1)} = \frac{1.46}{0.73} = 2$ m

Hence, the height of the pedestal is 2 m.

Page No 657:

Answer:

Let $A B$ be the unfinished tower, $A C$ be the raised tower and O be the point of observation.
We have:
$O A = 75$ m, ∠ $A O B = 30^{o}$ and ∠ $A O C = 60^{o}$
Let $A C = H$ m such that $B C = (H - h)$ m.

In ∆AOB, we have:
$\frac{A B}{O A} = \tan 30^{o} = \frac{1}{\sqrt{3}}$

⇒ $\frac{h}{75} = \frac{1}{\sqrt{3}}$

⇒ $h = \frac{75}{\sqrt{3}}$ m = $\frac{75 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = 25 \sqrt{3}$ m
In ∆ $A O C$ , we have:
$\frac{A C}{O A} = \tan 60^{o}$ $= \sqrt{3}$

⇒ $\frac{H}{75} = \sqrt{3}$
⇒ $H = 75 \sqrt{3}$ m

∴ Required height = $(H - h) = (75 \sqrt{3} - 25 \sqrt{3}) = 50 \sqrt{3}$ m = 86.6 m

Page No 657:

Answer:

Let $O X$ be the horizontal plane, $A D$ be the tower and $C D$ be the vertical flagpole.
We have:
$A B = 9$ m, ∠ $D B A = 30^{o}$ and ∠ $C B A = 60^{o}$
Let:
$A D = h$ m and $C D = x$ m

In the right ∆ $A B D$ , we have:
$\frac{A D}{A B} = \tan 30^{o} = \frac{1}{\sqrt{3}}$
⇒ $\frac{h}{9} = \frac{1}{\sqrt{3}}$
⇒ $h = \frac{9}{\sqrt{3}} = 5.19$ m
Now, in the right ∆ $A B C$ , we have:
$\frac{A C}{B A} = \tan 60^{o} = \sqrt{3}$

⇒ $\frac{h + x}{9} = \sqrt{3}$
⇒ $h + x = 9 \sqrt{3}$

By putting $h = \frac{9}{\sqrt{3}}$ in the above equation, we get:
$\frac{9}{\sqrt{3}} + x = 9 \sqrt{3}$
⇒ $x = 9 \sqrt{3} - \frac{9}{\sqrt{3}}$
⇒ $x = \frac{27 - 9}{\sqrt{3}} = \frac{18}{\sqrt{3}} = \frac{18}{1.73} = 10.4$

Thus, we have:
Height of the flagpole = 10.4 m
Height of the tower = 5.19 m

Page No 657:

Answer:

Let AB and CD be the equal poles; and BD be the width of the road.

We have,
$∠ AOB = 60 ° and ∠ COD = 60 ° In ∆ AOB, \tan 60 ° = \frac{AB}{BO} \Rightarrow \sqrt{3} = \frac{AB}{BO} \Rightarrow BO = \frac{AB}{\sqrt{3}} Also, in ∆ COD, \tan 30 ° = \frac{CD}{DO} \Rightarrow \frac{1}{\sqrt{3}} = \frac{CD}{DO} \Rightarrow DO = \sqrt{3} CD As, BD = 80 \Rightarrow BO + DO = 80 \Rightarrow \frac{AB}{\sqrt{3}} + \sqrt{3} CD = 80 \Rightarrow \frac{AB}{\sqrt{3}} + \sqrt{3} AB = 80 (Given : AB = CD) \Rightarrow AB (\frac{1}{\sqrt{3}} + \sqrt{3}) = 80 \Rightarrow AB (\frac{1 + 3}{\sqrt{3}}) = 80 \Rightarrow AB (\frac{4}{\sqrt{3}}) = 80 \Rightarrow AB = \frac{80 \sqrt{3}}{4} \Rightarrow AB = 20 \sqrt{3} m Also, BO = \frac{AB}{\sqrt{3}} = \frac{20 \sqrt{3}}{\sqrt{3}} = 20 m So, DO = 80 - 20 = 60 m$

Hence, the height of each pole is 20 $\sqrt{3}$ m and point P is at a distance of 20 m from left pole and 60 m from right pole.

Page No 658:

Answer:

Let $C D$ be the tower and $A and B$ be the positions of the two men standing on the opposite sides. Thus, we have:
∠ $D A C = 30^{o}$ , ∠ $D B C = 45^{o}$ and $C D = 50$ m
Let $A B = x$ m and $B C = y$ m such that $A C = (x - y)$ m.

In the right ∆ $D B C$ , we have:
$\frac{C D}{B C} = \tan 45^{o} = 1$

⇒ $\frac{50}{y} = 1$
⇒ $y = 50$ m

In the right ∆ $A C D$ , we have:
$\frac{C D}{A C} = \tan 30^{o} = \frac{1}{\sqrt{3}}$

⇒ $\frac{50}{(x - y)} = \frac{1}{\sqrt{3}}$
⇒ $x - y = 50 \sqrt{3}$
On putting $y = 50$ in the above equation, we get:
$x - 50 = 50 \sqrt{3}$
⇒ $x = 50 + 50 \sqrt{3} = 50 (\sqrt{3} + 1) = 136.6$ m

∴ Distance between the two men = $A B = x = 136.6$ m

Page No 658:

Answer:

Let PQ be the tower.

We have,

$PQ = 100 m, ∠ PAQ = 30 ° and ∠ PBQ = 45 ° In ∆ APQ, \tan 30 ° = \frac{PQ}{AP} \Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{AP} \Rightarrow AP = 100 \sqrt{3} m Als, in ∆ BPQ, \tan 45 ° = \frac{PQ}{BP} \Rightarrow 1 = \frac{100}{BP} \Rightarrow BP = 100 m Now, AB = AP + BP = 100 \sqrt{3} + 100 = 100 (\sqrt{3} + 1) = 100 \times (1.73 + 1) = 100 \times 2.73 = 273 m$

So, the distance between the cars is 273 m.

Page No 658:

Answer:

Let PQ be the tower.

$We have, ∠ PBQ = 60 ° and ∠ PAQ = 30 ° Let PQ = h, AB = x and BQ = y In ∆ APQ, \tan 30 ° = \frac{PQ}{AQ} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x + y} \Rightarrow x + y = h \sqrt{3} . . . . . (i) Also, in ∆ BPQ, \tan 60 ° = \frac{PQ}{BQ} \Rightarrow \sqrt{3} = \frac{h}{y} \Rightarrow h = y \sqrt{3} . . . . . (ii) Substituting h = y \sqrt{3} in (i), we get x + y = \sqrt{3} (y \sqrt{3}) \Rightarrow x + y = 3 y \Rightarrow 3 y - y = x \Rightarrow 2 y = x \Rightarrow y = \frac{x}{2} As, speed of the car from A to B = \frac{AB}{6} = \frac{x}{6} units / \sec So, the time taken to reach the foot of the tower i . e . Q from B = \frac{BQ}{speed} = \frac{y}{(\frac{x}{6})} = \frac{(\frac{x}{2})}{(\frac{x}{6})} = \frac{6}{2} = 3 \sec$

So, the time taken to reach the foot of the tower from the given point is 3 seconds.

Page No 658:

Answer:

Let PQ=h m be the height of the TV tower and BQ=x m be the width of the canal.

We have,

$AB = 20 m, ∠ PAQ = 30 °, ∠ PBQ = 60 °, BQ = x and PQ = h In ∆ PBQ, \tan 60 ° = \frac{PQ}{BQ} \Rightarrow \sqrt{3} = \frac{h}{x} \Rightarrow h = x \sqrt{3} . . . . . (i) Again, in ∆ APQ, \tan 30 ° = \frac{PQ}{AQ} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{AB + BQ} \Rightarrow \frac{1}{\sqrt{3}} = \frac{x \sqrt{3}}{20 + x} [Using (i)] \Rightarrow 3 x = 20 + x \Rightarrow 3 x - x = 20 \Rightarrow 2 x = 20 \Rightarrow x = \frac{20}{2} \Rightarrow x = 10 m Substituting x = 10 in (i), we get h = 10 \sqrt{3} m$

So, the height of the TV tower is $10 \sqrt{3} m$ and the width of the canal is 10 m.

Page No 658:

Answer:

Let AB be the building and PQ be the tower.

We have,

$PQ = 60 m, ∠ APB = 30 °, ∠ PAQ = 60 ° In ∆ APQ, \tan 60 ° = \frac{PQ}{AP} \Rightarrow \sqrt{3} = \frac{60}{AP} \Rightarrow AP = \frac{60}{\sqrt{3}} \Rightarrow AP = \frac{60 \sqrt{3}}{3} \Rightarrow AP = 20 \sqrt{3} m Now, in ∆ ABP, \tan 30 ° = \frac{AB}{AP} \Rightarrow \frac{1}{\sqrt{3}} = \frac{AB}{20 \sqrt{3}} \Rightarrow AB = \frac{20 \sqrt{3}}{\sqrt{3}} ∴ AB = 20 m$

So, the height of the building is 20 m.

Page No 658:

Answer:

Let $D E$ be the first tower and $A B$ be the second tower.
Now, $A B = 90$ m and $A D = 60$ m such that $C E = 60$ m and ∠ $B E C = 30^{o}$ .
Let $D E = h$ m such that $A C = h$ m and $B C = (90 - h)$ m.

In the right ∆ $B C E$ , we have:
$\frac{B C}{C E} = \tan 30^{o} = \frac{1}{\sqrt{3}}$
⇒ $\frac{(90 - h)}{60} = \frac{1}{\sqrt{3}}$
⇒ $(90 - h) \sqrt{3} = 60$
⇒ $h \sqrt{3} = 90 \sqrt{3} - 60$
⇒ $h = 90 - \frac{60}{\sqrt{3}}$ = $90 - 34.64 = 55.36$ m
∴ Height of the first tower = $D E = h = 55.36$ m

Page No 658:

Answer:

Let PQ be the chimney and AB be the tower.

We have,

$AB = 40 m, ∠ APB = 30 ° and ∠ PAQ = 60 ° In ∆ ABP, \tan 30 ° = \frac{AB}{AP} \Rightarrow \frac{1}{\sqrt{3}} = \frac{40}{AP} \Rightarrow AP = 40 \sqrt{3} m Now, in ∆ APQ, \tan 60 ° = \frac{PQ}{AP} \Rightarrow \sqrt{3} = \frac{PQ}{40 \sqrt{3}} ∴ PQ = 120 m$

So, the height of the chimney is 120 m.

Hence, the height of the chimney meets the pollution norms.

In this question, management of air pollution has been shown.
a

Page No 658:

Answer:

Let AB be the 7-m high building and CD be the cable tower.

We have,

$AB = 7 m, ∠ CAE = 60 °, ∠ DAE = ∠ ADB = 45 ° Also, DE = AB = 7 m In ∆ ABD, \tan 45 ° = \frac{AB}{BD} \Rightarrow 1 = \frac{7}{BD} \Rightarrow BD = 7 m So, AE = BD = 7 m Also, in ∆ ACE, \tan 60 ° = \frac{CE}{AE} \Rightarrow \sqrt{3} = \frac{CE}{7} \Rightarrow CE = 7 \sqrt{3} m Now, CD = CE + DE = 7 \sqrt{3} + 7 = 7 (\sqrt{3} + 1) m = 7 (1.732 + 1) = 7 (2.732) = 19.124 \approx 19.12 m$

So, the height of the tower is 19.12 m.

Page No 659:

Answer:

Let PQ be the tower.

We have,

$AB = 20 m, ∠ PAQ = 30 ° and ∠ PBQ = 60 ° Let BQ = x and PQ = h In ∆ PBQ, \tan 60 ° = \frac{PQ}{BQ} \Rightarrow \sqrt{3} = \frac{h}{x} \Rightarrow h = x \sqrt{3} . . . . . (i) Also, in ∆ APQ, \tan 30 ° = \frac{PQ}{AQ} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{AB + BQ} \Rightarrow \frac{1}{\sqrt{3}} = \frac{x \sqrt{3}}{20 + x} [Using (i)] \Rightarrow 20 + x = 3 x \Rightarrow 3 x - x = 20 \Rightarrow 2 x = 20 \Rightarrow x = \frac{20}{2} \Rightarrow x = 10 m From (i), h = 10 \sqrt{3} = 10 \times 1.732 = 17.32 m Also, AQ = AB + BQ = 20 + 10 = 30 m$

So, the height of the tower is 17.32 m and its distance from the point A is 30 m.

Page No 659:

Answer:

Let PQ be the tower.

We have,

$AB = 10 m, ∠ MAP = 30 ° and ∠ PBQ = 60 ° Also, MQ = AB = 10 m Let BQ = x and PQ = h So, AM = BQ = x and PM = PQ - MQ = h - 10 In ∆ BPQ, \tan 60 ° = \frac{PQ}{BQ} \Rightarrow \sqrt{3} = \frac{h}{x} \Rightarrow x = \frac{h}{\sqrt{3}} . . . . . (i) Now, in ∆ AMP, \tan 30 ° = \frac{PM}{AM} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h - 10}{x} \Rightarrow h \sqrt{3} - 10 \sqrt{3} = x \Rightarrow h \sqrt{3} - 10 \sqrt{3} = \frac{h}{\sqrt{3}} [Using (i)] \Rightarrow 3 h - 30 = h \Rightarrow 3 h - h = 30 \Rightarrow 2 h = 30 \Rightarrow h = \frac{30}{2} ∴ h = 15 m$

So, the height of the tower is 15 m.

Page No 659:

Answer:

Let AD be the tower and BC be the cliff.

We have,

$BC = 60 \sqrt{3} m, ∠ CDE = 45 ° and ∠ BAC = 60 ° Let AD = h \Rightarrow BE = AD = h \Rightarrow CE = BC - BE = 60 \sqrt{3} - h In ∆ CDE, \tan 45 ° = \frac{CE}{DE} \Rightarrow 1 = \frac{60 \sqrt{3} - h}{DE} \Rightarrow DE = 60 \sqrt{3} - h \Rightarrow AB = DE = 60 \sqrt{3} - h . . . . . (i) Now, in ∆ ABC, \tan 60 ° = \frac{BC}{AB} \Rightarrow \sqrt{3} = \frac{60 \sqrt{3}}{60 \sqrt{3} - h} [Using (i)] \Rightarrow 180 - h \sqrt{3} = 60 \sqrt{3} \Rightarrow h \sqrt{3} = 180 - 60 \sqrt{3} \Rightarrow h = \frac{180 - 60 \sqrt{3}}{\sqrt{3}} \Rightarrow h = \frac{180 - 60 \sqrt{3}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \Rightarrow h = \frac{180 \sqrt{3} - 180}{3} \Rightarrow h = \frac{180 (\sqrt{3} - 1)}{3} ∴ h = 60 (\sqrt{3} - 1) = 60 (1.732 - 1) = 60 (0.732) Also, h = 43.92 m$

So, the height of the tower is 43.92 m.

Page No 659:

Answer:

Let $A B$ be the deck of the ship above the water level and $D E$ be the cliff.
Now,
$A B = 16$ m such that $C D = 16$ m and ∠ $B D A = 30^{o}$ and ∠ $E B C = 60^{o}$ .
If AD = x m and $D E = h$ m, then $C E = (h - 16)$ m.

In the right ∆ $B A D$ , we have:
$\frac{A B}{A D} = \tan 30^{o} = \frac{1}{\sqrt{3}}$

⇒ $\frac{16}{x} = \frac{1}{\sqrt{3}}$

⇒ $x = 16 \sqrt{3}$ $= 27.68$ m

In the right ∆ $E B C$ , we have:
$\frac{E C}{B C} = \tan 60^{o} = \sqrt{3}$

⇒ $\frac{(h - 16)}{x} = \sqrt{3}$
⇒ $h - 16 = x \sqrt{3}$
⇒ $h - 16 = 16 \sqrt{3} \times \sqrt{3} = 48$ [∵ $x = 16 \sqrt{3}$ ]
⇒ $h = 48 + 16 = 64$ m

∴ Distance of the cliff from the deck of the ship = $A D = x = 27.68$ m
And,
Height of the cliff = $D E = h = 64$ m

Page No 659:

Answer:

We have,

$XY = 40 m, ∠ PXQ = 60 ° and ∠ MYQ = 45 ° Let PQ = h Also, MP = XY = 40 m, MQ = PQ - MP = h - 40 In ∆ MYQ, \tan 45 ° = \frac{MQ}{MY} \Rightarrow 1 = \frac{h - 40}{MY} \Rightarrow MY = h - 40 \Rightarrow PX = MY = h - 40 . . . . . (i) Now, in ∆ MXQ, \tan 60 ° = \frac{PQ}{PX} \Rightarrow \sqrt{3} = \frac{h}{h - 40} [From (i)] \Rightarrow h \sqrt{3} - 40 \sqrt{3} = h \Rightarrow h \sqrt{3} - h = 40 \sqrt{3} \Rightarrow h (\sqrt{3} - 1) = 40 \sqrt{3} \Rightarrow h = \frac{40 \sqrt{3}}{(\sqrt{3} - 1)} \Rightarrow h = \frac{40 \sqrt{3}}{(\sqrt{3} - 1)} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)} \Rightarrow h = \frac{40 \sqrt{3} (\sqrt{3} + 1)}{(3 - 1)} \Rightarrow h = \frac{40 \sqrt{3} (\sqrt{3} + 1)}{2} \Rightarrow h = 20 \sqrt{3} (\sqrt{3} + 1) \Rightarrow h = 60 + 20 \sqrt{3} \Rightarrow h = 60 + 20 \times 1.73 \Rightarrow h = 60 + 34.6 ∴ h = 94.6 m$

So, the height of the tower PQ is 94.6 m.

Page No 659:

Answer:

Let the height of flying of the aeroplane be PQ = BC and point A be the point of observation.

We have,

$PQ = BC = 2500 m, ∠ PAQ = 45 ° and ∠ BAC = 30 ° In ∆ PAQ, \tan 45 ° = \frac{PQ}{AQ} \Rightarrow 1 = \frac{2500}{AQ} \Rightarrow AQ = 2500 m Also, in ∆ ABC, \tan 30 ° = \frac{BC}{AC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{2500}{AC} \Rightarrow AC = 2500 \sqrt{3} m Now, QC = AC - AQ = 2500 \sqrt{3} - 2500 = 2500 (\sqrt{3} - 1) m = 2500 (1.732 - 1) = 2500 (0.732) = 1830 m \Rightarrow PB = QC = 1830 m So, the speed of the aeroplane = \frac{PB}{15} = \frac{1830}{15} = 122 m / s = 122 \times \frac{3600}{1000} km / h = 439.2 km / h$

So, the speed of the aeroplane is 122 m/s or 439.2 km/h.

Page No 659:

Answer:

Let $A B$ be the tower.
We have:
$C D = 150$ m, ∠ $A C B = 30^{o}$ and ∠ $A D B = 60^{o}$
Let:
$A B = h$ m and $B D = x$ m

In the right ∆ $A B D$ , we have:
$\frac{A B}{A D} = \tan 60^{o} = \sqrt{3}$

⇒ $\frac{h}{x} = \sqrt{3}$
⇒ $x = \frac{h}{\sqrt{3}}$
Now, in the right ∆ $A C B$ , we have:
$\frac{A B}{B C} = \tan 30^{o} = \frac{1}{\sqrt{3}}$

⇒ $\frac{h}{x + 150} = \frac{1}{\sqrt{3}}$
⇒ $\sqrt{3} h = x + 150$

On putting $x = \frac{h}{\sqrt{3}}$ in the above equation, we get:
$\sqrt{3} h = \frac{h}{\sqrt{3}} + 150$
⇒ $3 h = h + 150 \sqrt{3}$
⇒ $2 h = 150 \sqrt{3}$
⇒ $h = \frac{150 \sqrt{3}}{2} = 75 \sqrt{3} = 75 \times 1.732 = 129.9$ m
Hence, the height of the tower is 129.9 m.

Page No 659:

Answer:

Let $O A$ be the lighthouse and B and C be the two positions of the ship.
Thus, we have:
$O A = 100$ m, ∠ $O B A = 30^{o}$ and ∠ $O C A = 60^{o}$

Let:
$O C = x$ m and $B C = y$ m
In the right ∆ $O A C$ , we have:
$\frac{O A}{O C} = \tan 60^{o} = \sqrt{3}$

⇒ $\frac{100}{x} = \sqrt{3}$
⇒ $x = \frac{100}{\sqrt{3}}$ m
Now, in the right ∆ $O B A$ , we have:
$\frac{O A}{O B} = \tan 30^{o} = \frac{1}{\sqrt{3}}$

⇒ $\frac{100}{x + y} = \frac{1}{\sqrt{3}}$
⇒ $x + y = 100 \sqrt{3}$

On putting $x = \frac{100}{\sqrt{3}}$ in the above equation, we get:
$y = 100 \sqrt{3} - \frac{100}{\sqrt{3}} = \frac{300 - 100}{\sqrt{3}} = \frac{200}{\sqrt{3}} = 115.47$ m

∴ Distance travelled by the ship during the period of observation = $B C = y = 115.47$ m

Page No 659:

Answer:

Let $A$ and $B$ be two points on the banks on the opposite side of the river and $P$ be the point on the bridge at a height of 2.5 m.
Thus, we have:
$D P = 2.5$ m, ∠PAD $= 30^{o}$ and ∠ $P B D = 45^{o}$
In the right ∆ $A P D$ , we have:
$\frac{D P}{A D} = \tan 30^{o} = \frac{1}{\sqrt{3}}$
⇒ $\frac{2.5}{A D} = \frac{1}{\sqrt{3}}$
⇒ $A D = 2.5 \sqrt{3}$ m
In the right ∆ $P D B$ , we have:
$\frac{D P}{B D} = \tan 45^{o} = 1$
⇒ $\frac{2.5}{B D} = 1$
⇒ $B D = 2.5$ m

∴ Width of the river = $A B = (A D + B D) = (2.5 \sqrt{3} + 2.5) = 6.83$ m

Page No 659:

Answer:

Let $A B$ be the tower and $C and D$ be two points such that $A C = 4$ m and $A D = 9$ m.
Let:
$A B = h$ m, ∠ $B C A = θ$ and ∠ $B D A = 90^{°} - θ$

In the right ∆BCA, we have:
$\tan θ = \frac{A B}{A C} \Rightarrow \tan θ = \frac{h}{4} . . . (1)$
In the right ∆BDA, we have:
$\tan (90 ° - θ) = \frac{A B}{A D} \Rightarrow \cot θ = \frac{h}{9} [\tan (90 ° - θ) = \cot θ] \Rightarrow \frac{1}{\tan θ} = \frac{h}{9} . . . (2) [\cot θ = \frac{1}{\tan θ}]$
Multiplying equations (1) and (2), we get:
$\tan θ \times \frac{1}{\tan θ} = \frac{h}{4} \times \frac{h}{9} \Rightarrow 1 = \frac{h^{2}}{36}$
⇒ 36 = h²
⇒ h = ±6

Height of a tower cannot be negative.
∴ Height of the tower = 6 m

Page No 660:

Answer:

Let AB and CD be the two opposite walls of the room and the foot of the ladder be fixed at the point O on the ground.

We have,

$AO = CO = 6 m, ∠ AOB = 60 ° and ∠ COD = 45 ° In ∆ ABO, \cos 60 ° = \frac{BO}{AO} \Rightarrow \frac{1}{2} = \frac{BO}{6} \Rightarrow BO = \frac{6}{2} \Rightarrow BO = 3 m Also, in ∆ CDO, \cos 45 ° = \frac{DO}{CO} \Rightarrow \frac{1}{\sqrt{2}} = \frac{DO}{6} \Rightarrow DO = \frac{6}{\sqrt{2}} \Rightarrow DO = \frac{6}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \Rightarrow DO = \frac{6 \sqrt{2}}{2} \Rightarrow DO = 3 \sqrt{2} m Now, the distance between two walls of the room = BD = BO + DO = 3 + 3 \sqrt{2} = 3 (1 + \sqrt{2}) = 3 (1 + 1.414) = 3 (2.414) = 7.242 \approx 7.24 m$

So, the distance between two walls of the room is 7.24 m.

Page No 660:

Answer:

Let OP be the tower and points A and B be the positions of the cars.

We have,

$AB = 100 m, ∠ OAP = 60 ° and ∠ OBP = 45 ° Let OP = h In ∆ AOP, \tan 60 ° = \frac{OP}{OA} \Rightarrow \sqrt{3} = \frac{h}{OA} \Rightarrow OA = \frac{h}{\sqrt{3}} Also, in ∆ BOP, \tan 45 ° = \frac{OP}{OB} \Rightarrow 1 = \frac{h}{OB} \Rightarrow OB = h Now, OB - OA = 100 \Rightarrow h - \frac{h}{\sqrt{3}} = 100 \Rightarrow \frac{h \sqrt{3} - h}{\sqrt{3}} = 100 \Rightarrow \frac{h (\sqrt{3} - 1)}{\sqrt{3}} = 100 \Rightarrow h = \frac{100 \sqrt{3}}{(\sqrt{3} - 1)} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)} \Rightarrow h = \frac{100 \sqrt{3} (\sqrt{3} + 1)}{(3 - 1)} \Rightarrow h = \frac{100 (3 + \sqrt{3})}{2} \Rightarrow h = 50 (3 + 1.732) \Rightarrow h = 50 (4.732) ∴ h = 236.6 m$

So, the height of the tower is 236.6 m.

Disclaimer: The answer given in the texbook is incorrect. The same has been rectified above.

Page No 660:

Answer:

Let AC be the pole and BD be the ladder.

We have,

$AC = 4 m, AB = 1 m and ∠ BDC = 60 ° And, BC = AC - AB = 4 - 1 = 3 m In ∆ BDC, \sin 60 ° = \frac{BC}{BD} \Rightarrow \frac{\sqrt{3}}{2} = \frac{3}{BD} \Rightarrow BD = \frac{3 \times 2}{\sqrt{3}} \Rightarrow BD = 2 \sqrt{3} \Rightarrow BD = 2 \times 1.73 ∴ BD = 3.46 m$

So, he should use 3.46 m long ladder to reach the required position.

Page No 660:

Answer:

We have,

$AB = 60 m, ∠ ACE = 30 ° and ∠ ADB = 60 ° Let BD = CE = x and CD = BE = y \Rightarrow AE = AB - BE = 60 - y In ∆ ACE, \tan 30 ° = \frac{AE}{CE} \Rightarrow \frac{1}{\sqrt{3}} = \frac{60 - y}{x} \Rightarrow x = 60 \sqrt{3} - y \sqrt{3} . . . . . (i) Also, in ∆ ABD, \tan 60 ° = \frac{AB}{BD} \Rightarrow \sqrt{3} = \frac{60}{x} \Rightarrow x = \frac{60}{\sqrt{3}} \Rightarrow x = \frac{60}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \Rightarrow x = \frac{60 \sqrt{3}}{3} \Rightarrow x = 20 \sqrt{3} Substituting x = 20 \sqrt{3} in (i), we get 20 \sqrt{3} = 60 \sqrt{3} - y \sqrt{3} \Rightarrow y \sqrt{3} = 60 \sqrt{3} - 20 \sqrt{3} \Rightarrow y \sqrt{3} = 40 \sqrt{3} \Rightarrow y = \frac{40 \sqrt{3}}{\sqrt{3}} \Rightarrow y = 40 m (i) the horizontal distance between AB and CD = BD = x = 20 \sqrt{3} = 20 \times 1.732 = 34.64 m (ii) the height of the lamp post = CD = y = 40 m (iii) the difference between the heights of the building and the lamp post = AB - CD = 60 - 40 = 20 m$

Page No 660:

Answer:

Suppose AB be the tower of height h meters. Let C be the initial position of the car and let after 12 minutes the car be at D. It is given that the angles of depression at C and D are 30º and 45º respectively.
Let the speed of the car be v meter per minute. Then,
CD = distance travelled by the car in 12 minutes
CD = 12v meters

Suppose the car takes t minutes to reach the tower AB from D. Then DA = vt meters.
$In ∆ DAB, we have \tan 45 ° = \frac{AB}{AD} \Rightarrow 1 = \frac{h}{v t} \Rightarrow h = v t . . . . . (i) In ∆ CAB, we have \tan 30 ° = \frac{AB}{AC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{v t + 12 v} \Rightarrow \sqrt{3} h = v t + 12 v . . . . . (i i)$
Substituting the value of h from equation (i) in equation (ii), we get
$\Rightarrow \sqrt{3} t = t + 12 \Rightarrow t = \frac{12}{\sqrt{3} - 1} = \frac{12}{(\sqrt{3} - 1)} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)} = 6 (\sqrt{3} + 1) \min$

Page No 660:

Answer:

Let CD be the height of the aeroplane above the river at some instant. Suppose A and B be two points on both banks of the river in opposite directions.

Height of the aeroplane above the river, CD = 300 m
Now,
$∠$ CAD = $∠$ ADX = 60º (Alternate angles)
$∠$ CBD = $∠$ BDY = 45º (Alternate angles)
In right ∆ACD,
$\tan 60 ° = \frac{CD}{AC} \Rightarrow \sqrt{3} = \frac{300}{AC} \Rightarrow AC = \frac{300}{\sqrt{3}} = 100 \sqrt{3} m$
In right ∆BCD,
$\tan 45 ° = \frac{CD}{BC} \Rightarrow 1 = \frac{300}{BC} \Rightarrow BC = 300 m$
∴ Width of the river, AB
= BC + AC
$= 300 + 100 \sqrt{3} = 300 + 100 \times 1.73 = 300 + 173 = 473 m$
Thus, the width of the river is 473 m.

Page No 660:

Answer:

Let BC be the 20 m high building and AB be the communication tower of height h fixed on top of the building. Let D be a point on ground such that CD = x m and angles of elevation made from this point to top and bottom of tower are $45 °$ and $60 °$ .
In $△ BCD : \tan 45 ° = \frac{BC}{CD} = \frac{20}{x}$
$\Rightarrow 1 = \frac{20}{x} \Rightarrow x = 20 m .$

Also, in $△ ACD : \tan 60 ° = \frac{AC}{CD} = \frac{20 + h}{x}$
$\Rightarrow \sqrt{3} = \frac{20 + h}{x} \Rightarrow \sqrt{3} = \frac{20 + h}{20} \Rightarrow 20 + h = 20 \sqrt{3} \Rightarrow h = 20 \sqrt{3} - 20 \Rightarrow h = 20 (\sqrt{3} - 1) = 20 (1.73 - 1) = 20 \times 0.73 \Rightarrow h = 14.64 m .$

Page No 660:

Answer:

Let PQ be the hill of height h km. Let R and S be two consecutive kilometre stones, so the distance between them is 1 km.
Let QR = x km.

$In ∆ PQR, \tan 45 ° = \frac{PQ}{QR} \Rightarrow 1 = \frac{h}{x} \Rightarrow h = x . . . (i)$

$In ∆ PQS, \tan 30 ° = \frac{PQ}{QS} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x + 1} \Rightarrow \sqrt{3} h = x + 1 . . . (ii)$
From equation (i) and (ii) we get,
$\sqrt{3} h = h + 1 \Rightarrow h (\sqrt{3} - 1) = 1 \Rightarrow h = \frac{1}{\sqrt{3} - 1} = \frac{\sqrt{3} + 1}{(\sqrt{3} - 1) (\sqrt{3} + 1)} \Rightarrow h = \frac{\sqrt{3} + 1}{2} = \frac{2.73}{2} = 1.365 km$
Hence, the height of the hill is 1.365 km.

Page No 661:

Answer:

Let AB be the vertically standing pole of height h units and CB be the length of its shadow of s units.
Since, the ratio of length of pole and its shadow at some time of day is given to be $\sqrt{3} : 1 .$
$\Rightarrow \frac{AB}{BC} = \frac{\sqrt{3}}{1} .$
$∴$ In $△ ABC :$
$\tan (θ) = \frac{AB}{BC} = \frac{\sqrt{3}}{1} = \sqrt{3} \Rightarrow \tan (θ) = \tan 60 ° \Rightarrow θ = 60 ° .$

Page No 661:

Answer:

Let AB be the light house. Suppose C and D be the two positions of the boat.

Here, AB = 100 m.

Let the speed of the boat be v m/min.

So,

CD = v m/min × 2 min = 2v m [Distance = Speed × Time]

In right ∆ABC,

$\tan 60 ° = \frac{AB}{BC} \Rightarrow \sqrt{3} = \frac{100 m}{BC} \Rightarrow BC = \frac{100}{\sqrt{3}} = \frac{100 \sqrt{3}}{3} m . . . . . (1)$

In right ∆ABD,

$\tan 30 ° = \frac{AB}{BD} \Rightarrow \frac{1}{\sqrt{3}} = \frac{100 m}{BC + CD} \Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{\frac{100 \sqrt{3}}{3} + 2 v} [From (1)]$
$\Rightarrow 2 v + \frac{100 \sqrt{3}}{3} = 100 \sqrt{3} \Rightarrow 2 v = 100 \sqrt{3} - \frac{100 \sqrt{3}}{3} \Rightarrow 2 v = 100 \sqrt{3} (1 - \frac{1}{3}) \Rightarrow 2 v = 100 \sqrt{3} \times \frac{2}{3}$
$\Rightarrow v = \frac{100 \times 1.732}{3} = 57.73 m / \min$

Thus, the speed boat is 57.73 m/min.

Page No 661:

Answer:

Let AB be the building and CD be the tower.

Draw AE ⊥ CD.

Suppose the height of the tower be h m.

Here, AB = 8 m

∠DAE = ∠ADX = 30º        (Alternate angles)

∠DBC = ∠BDX = 45º        (Alternate angles)

CE = AB = 8 m

∴ DE = CD − CE = (h − 8) m

Distance between the building and tower = BC

In right ∆BCD,

$\tan 45 ° = \frac{CD}{BC} \Rightarrow 1 = \frac{h}{BC} \Rightarrow BC = h m$

In right ∆AED,

$\tan 30 ° = \frac{DE}{AE} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h - 8}{h} (AE = BC) \Rightarrow \sqrt{3} h - 8 \sqrt{3} = h$
$\Rightarrow \sqrt{3} h - h = 8 \sqrt{3} \Rightarrow (\sqrt{3} - 1) h = 8 \sqrt{3} \Rightarrow h = \frac{8 \sqrt{3}}{\sqrt{3} - 1} \Rightarrow h = \frac{8 \sqrt{3} (\sqrt{3} + 1)}{(\sqrt{3} - 1) \times (\sqrt{3} + 1)}$
$\Rightarrow h = \frac{8 \sqrt{3} (\sqrt{3} + 1)}{2} [(a - b) (a + b) = a^{2} - b^{2}] \Rightarrow h = 4 \sqrt{3} (\sqrt{3} + 1) \Rightarrow h = (12 + 4 \sqrt{3}) m$

∴ Height of the tower = $(12 + 4 \sqrt{3}) m$

Also,

Distance between the tower and building = BC = $(12 + 4 \sqrt{3}) m$                  (BC = h m)

Thus, the the height of the tower is $(12 + 4 \sqrt{3}) m$ and the distance between the tower and the building is $(12 + 4 \sqrt{3}) m$ .

Page No 671:

Answer:

Let AB represents the vertical pole and BC represents the shadow on the ground and θ represents angle of elevation the sun.

$In ∆ ABC, \tan θ = \frac{AB}{BC} \Rightarrow \tan θ = \frac{x}{x} (As, the height of the pole, AB = the length of the shadow, BC = x) \Rightarrow \tan θ = 1 \Rightarrow \tan θ = \tan 45 ° ∴ θ = 45 °$

Hence, the correct answer is option (c).

Page No 671:

Answer:

Here, AO be the pole; BO be its shadow and $θ$ be the angle of elevation of the sun.

$Let BO = x Then, AO = x \sqrt{3} In ∆ AOB, \tan θ = \frac{AO}{BO} \Rightarrow \tan θ = \frac{x \sqrt{3}}{x} \Rightarrow \tan θ = \sqrt{3} \Rightarrow \tan θ = \tan 60 ° ∴ θ = 60 °$

Hence, the correct answer is option (c).

Page No 671:

Answer:

(b) 30°

Let $A B$ be the pole and $B C$ be its shadow.

Let $A B = h$ and $B C = x$ such that $x = \sqrt{3} h$ (given) and $θ$ be the angle of elevation.
From ∆ $A B C$ , we have:
$\frac{A B}{B C} = \tan θ$

⇒ $\frac{h}{x} = \frac{h}{\sqrt{3} h} = \tan θ$
⇒ $\tan θ = \frac{1}{\sqrt{3}}$
⇒ $θ = 30^{o}$

Hence, the angle of elevation is $30^{o}$ .

Page No 671:

Answer:

Let AB be the pole, BC be its shadow and $θ$ be the sun's elevation.

$We have, AB = 12 m and BC = 4 \sqrt{3} m In ∆ ABC, \tan θ = \frac{AB}{BC} \Rightarrow \tan θ = \frac{12}{4 \sqrt{3}} \Rightarrow \tan θ = \frac{3}{\sqrt{3}} \Rightarrow \tan θ = \frac{3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \Rightarrow \tan θ = \frac{3 \sqrt{3}}{3} \Rightarrow \tan θ = \sqrt{3} \Rightarrow \tan θ = \tan 60 ° ∴ θ = 60 °$

Hence, the correct answer is option (a).

Page No 671:

Answer:

Let AB be a stick and BC be its shadow; and PQ be the tree and QR be its shadow.

We have,

$AB = 5 m, BC = 2 m, PQ = 12.5 m In ∆ ABC, \tan θ = \frac{AB}{BC} \Rightarrow \tan θ = \frac{5}{2} . . . . . (i) Now, in ∆ PQR, \tan θ = \frac{PQ}{QR} \Rightarrow \frac{5}{2} = \frac{12.5}{QR} [Using (i)] \Rightarrow QR = \frac{12.5 \times 2}{5} = \frac{25}{5} ∴ QR = 5 m$

Hence, the correct answer is option (d).

Page No 672:

Answer:

Let AB be the wall and AC be the ladder.

We have,

BC = 2 m and ∠ ACB = 60 ° In ∆ ABC, \cos 60 ° = \frac{BC}{AC} \Rightarrow \frac{1}{2} = \frac{2}{AC} ∴ AC = 4 m

Hence, the correct answer is option (d).

Page No 672:

Answer:

Let AB be the wall and AC be the ladder.

We have,

$AC = 15 m and ∠ BAC = 60 ° In ∆ ABC, \cos 60 ° = \frac{AB}{AC} \Rightarrow \frac{1}{2} = \frac{AB}{15} ∴ AB = \frac{15}{2} m$

Hence, the correct answer is option (c).

Page No 672:

Answer:

Let AB be the tower and point C be the point of observation on the ground.

We have,

$BC = 30 m and ∠ ACB = 30 ° In ∆ ABC, \tan 30 ° = \frac{AB}{BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{AB}{30} \Rightarrow AB = \frac{30}{\sqrt{3}} \Rightarrow AB = \frac{30}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \Rightarrow AB = \frac{30 \sqrt{3}}{3} ∴ AB = 10 \sqrt{3} m$

Hence, the correct answer is option (b).

Page No 672:

Answer:

Let AB be the tower and point C be the position of the car.

We have,

$AB = 150 m and ∠ ACB = 30 ° In ∆ ABC, \tan 30 ° = \frac{AB}{BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{150}{BC} ∴ BC = 150 \sqrt{3} m$

Hence, the correct answer is option (b).

Page No 672:

Answer:

Let point A be the position of the kite and AC be its string.

We have,

$AB = 30 m and AC = 60 m Let ∠ ACB = θ In ∆ ABC, \sin θ = \frac{AB}{AC} \Rightarrow \sin θ = \frac{30}{60} \Rightarrow \sin θ = \frac{1}{2} \Rightarrow \sin θ = \sin 30 ° ∴ θ = 30 °$

Hence, the correct answer is option (b).

Page No 672:

Answer:

Let AB be the cliff and CD be the tower.

We have,

$AB = 20 m Also, CE = AB = 20 m Let ∠ ACB = ∠ CAE = ∠ DAE = θ In ∆ ABC, \tan θ = \frac{AB}{BC} \Rightarrow \tan θ = \frac{20}{BC} \Rightarrow \tan θ = \frac{20}{AE} (As, BC = AE) \Rightarrow AE = \frac{20}{\tan θ} . . . . . (i) Also, in ∆ ADE, \tan θ = \frac{DE}{AE} \Rightarrow \tan θ = \frac{DE}{(\frac{20}{\tan θ})} [Using (i)] \Rightarrow \tan θ = \frac{DE \times \tan θ}{20} \Rightarrow DE = \frac{20 \times \tan θ}{\tan θ} \Rightarrow DE = 20 m Now, CD = DE + CE = 20 + 20 ∴ CD = 40 m$

Hence, the correct answer is option (b).

Disclaimer: The answer given in the textbook is incorrect. The same has been rectified above.

Page No 672:

Answer:

Let AB be the lamp post; CD be the girl and DE be her shadow.

We have,

$CD = 1.5 m, AD = 3 m, DE = 4.5 m Let ∠ E = θ In ∆ CDE, \tan θ = \frac{CD}{DE} \Rightarrow \tan θ = \frac{1.5}{4.5} \Rightarrow \tan θ = \frac{1}{3} . . . . . (i) Now, in ∆ ABE, \tan θ = \frac{AB}{AE} \Rightarrow \frac{1}{3} = \frac{AB}{AD + DE} [Using (i)] \Rightarrow \frac{1}{3} = \frac{AB}{3 + 4.5} \Rightarrow AB = \frac{7.5}{3} ∴ AB = 2.5 m$

Hence, the correct answer is option (c).

Page No 672:

Answer:

Let CD = h be the height of the tower.

We have,

$AB = 2 x, ∠ DAC = 30 ° and ∠ DBC = 45 ° In ∆ BCD, \tan 45 ° = \frac{CD}{BC} \Rightarrow 1 = \frac{h}{BC} \Rightarrow BC = h Now, in ∆ ACD, \tan 30 ° = \frac{CD}{AC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{AB + BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{2 x + h} \Rightarrow 2 x + h = h \sqrt{3} \Rightarrow h \sqrt{3} - h = 2 x \Rightarrow h (\sqrt{3} - 1) = 2 x \Rightarrow h = \frac{2 x}{(\sqrt{3} - 1)} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)} \Rightarrow h = \frac{2 x (\sqrt{3} + 1)}{(3 - 1)} \Rightarrow h = \frac{2 x (\sqrt{3} + 1)}{2} ∴ h = x (\sqrt{3} + 1) m$

Hence, the correct answer is option (d).

Page No 672:

Answer:

Let AB be the rod and BC be its shadow; and $θ$ be the angle of elevation of the sun.

$We have, AB : BC = 1 : \sqrt{3} Let AB = x Then, BC = x \sqrt{3} In ∆ ABC, \tan θ = \frac{AB}{BC} \Rightarrow \tan θ = \frac{x}{x \sqrt{3}} \Rightarrow \tan θ = \frac{1}{\sqrt{3}} \Rightarrow \tan θ = \tan 30 ° ∴ θ = 30 °$

Hence, the correct answer is option (a).

Page No 673:

Answer:

Let AB be the pole and BC be its shadow.

We have,

$BC = 2 \sqrt{3} m and ∠ ACB = 60 ° In ∆ ABC, \tan 60 ° = \frac{AB}{BC} \Rightarrow \sqrt{3} = \frac{AB}{2 \sqrt{3}} ∴ AB = 6 m$

Hence, the correct answer is option (b).

Page No 673:

Answer:

Let the sun's altitude be $θ$ .

We have,

$AB = 20 m and BC = 20 \sqrt{3} m In ∆ ABC, \tan θ = \frac{AB}{BC} \Rightarrow \tan θ = \frac{20}{20 \sqrt{3}} \Rightarrow \tan θ = \frac{1}{\sqrt{3}} \Rightarrow \tan θ = \tan 30 ° ∴ θ = 30 °$

Hence, the correct answer is option (a).

Page No 673:

Answer:

Let AB and CD be the two towers such that AB = x and CD = y.

We have,

$∠ AEB = 30 °, ∠ CED = 60 ° and BE = DE In ∆ ABE, \tan 30 ° = \frac{AB}{BE} \Rightarrow \frac{1}{\sqrt{3}} = \frac{x}{BE} \Rightarrow BE = x \sqrt{3} Also, in ∆ CDE, \tan 60 ° = \frac{CD}{DE} \Rightarrow \sqrt{3} = \frac{y}{DE} \Rightarrow DE = \frac{y}{\sqrt{3}} As, BE = DE \Rightarrow x \sqrt{3} = \frac{y}{\sqrt{3}} \Rightarrow \frac{x}{y} = \frac{1}{\sqrt{3} \times \sqrt{3}} \Rightarrow \frac{x}{y} = \frac{1}{3} ∴ x : y = 1 : 3$

Hence, the correct answer is option (c).

Page No 673:

Answer:

(b) $10 \sqrt{3} m$
Let $A B$ be the tower and $O$ be the point of observation.
Also,
∠ $A O B = 30^{o}$ and $O B = 30$ m
Let:
$A B = h$ m

In ∆ $A O B$ , we have:
$\frac{A B}{O B} = \tan 30^{o} = \frac{1}{\sqrt{3}}$

⇒ $\frac{h}{30} = \frac{1}{\sqrt{3}}$
⇒ $h = \frac{30}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{30 \sqrt{3}}{3} = 10 \sqrt{3}$ m

Hence, the height of the tower is $10 \sqrt{3}$ m.

Page No 673:

Answer:

(a) $50 \sqrt{3} m$
Let $A B$ be the string of the kite and $A X$ be the horizontal line.
If $B C$ ⊥ $A X$ , then $A B = 100$ m and ∠ $B A C = 60^{o}$ .
Let:
$B C = h$ m

In the right ∆ $A C B$ , we have:
$\frac{B C}{A B} = \sin 60^{o} = \frac{\sqrt{3}}{2}$

⇒ $\frac{h}{100} = \frac{\sqrt{3}}{2}$
⇒ $h = \frac{100 \sqrt{3}}{2} = 50 \sqrt{3}$ m
Hence, the height of the kite is $50 \sqrt{3}$ m.

Page No 673:

Answer:

(b) $\sqrt{a b}$
Let $A B$ be the tower and $C$ and $D$ be the points of observation on $A C$ .
Let:
∠ $A C B = θ$ , ∠ $A D B = 90 - θ$ and $A B = h$ m
Thus, we have:
$A C = a, A D = b$ and $C D = a - b$

Now, in the right ∆ABC, we have:
$\tan θ = \frac{A B}{A C}$ ⇒ $\frac{h}{a} = \tan θ$ ...(i)
In the right ∆ABD, we have:
$\tan (90 - θ) = \frac{A B}{A D}$ ⇒ $c o t θ = \frac{h}{b}$ ...(ii)

On multiplying (i) and (ii), we have:
$\tan θ \times c o t θ = \frac{h}{a} \times \frac{h}{b}$
⇒ $\frac{h}{a} \times \frac{h}{b} = 1$ [ ∵ $\tan θ = \frac{1}{c o t θ}$ ]
⇒ $h^{2} = a b$
⇒ $h = \sqrt{a b}$ m

Hence, the height of the tower is $\sqrt{a b}$ m.

Page No 673:

Answer:

(b) $10 \sqrt{3} m$
Let $A B$ be the tower and $C$ and $D$ be the points of observation such that ∠ $B C D = 30^{o}$ , ∠ $B D A = 60^{o}$ , $C D = 20$ m and $A D = x$ m.

Now, in ∆ $A D B$ , we have:
$\frac{A B}{A D} = \tan 60^{o}$ $= \sqrt{3}$

⇒ $\frac{A B}{x} = \sqrt{3}$
⇒ $A B = \sqrt{3} x$

In ∆ $A C B$ , we have:
$\frac{A B}{A C} = \tan 30^{o} = \frac{1}{\sqrt{3}}$
$\frac{A B}{20 + x} = \frac{1}{\sqrt{3}}$ ⇒ $A B = \frac{20 + x}{\sqrt{3}}$
∴ $\sqrt{3} x = \frac{20 + x}{\sqrt{3}}$
⇒ $3 x = 20 + x$
⇒ $2 x = 20$ ⇒ $x = 10$
∴ Height of the tower AB = $\sqrt{3} x = 10 \sqrt{3}$ m

Page No 673:

Answer:

(c) $16 \sqrt{3} {cm}^{2}$
Let $A B C D$ be the rectangle in which ∠ $B A C = 30^{o}$ and $A C = 8$ cm.

In ∆ $B A C$ , we have:

$\frac{A B}{A C} = \cos 30^{o} = \frac{\sqrt{3}}{2}$

⇒ $\frac{A B}{8} = \frac{\sqrt{3}}{2}$
⇒ $A B = 8 \frac{\sqrt{3}}{2} = 4 \sqrt{3}$ m
Again,

$\frac{B C}{A C} = \sin 30^{o} = \frac{1}{2}$

⇒ $\frac{B C}{8} = \frac{1}{2}$
⇒ $B C = \frac{8}{2} = 4$ m
∴ Area of the rectangle = $(A B \times B C) = (4 \sqrt{3} \times 4) = 16 \sqrt{3}$ cm²

Page No 673:

Answer:

(b) $\frac{1}{2} (\sqrt{3} + 1) km$
Let $A B$ be the hill making angles of depression at points $C$ and $D$ such that ∠ $A D B = 45^{o}$ , ∠ $A C B = 30^{o}$ and $C D = 1$ km.
Let:
$A B = h$ km and $A D = x$ km

In ∆ $A D B$ , we have:
$\frac{A B}{A D} = \tan 45^{o} = 1$

⇒ $\frac{h}{x} = 1$ ⇒ $h = x$ ...(i)
In ∆ $A C B$ , we have:
$\frac{A B}{A C} = \tan 30^{o} = \frac{1}{\sqrt{3}}$

⇒ $\frac{h}{x + 1} = \frac{1}{\sqrt{3}}$ ...(ii)
On putting the value of $h$ taken from (i) in (ii), we get:
$\frac{h}{h + 1} = \frac{1}{\sqrt{3}}$
⇒ $\sqrt{3} h = h + 1$
⇒ $(\sqrt{3} - 1) h = 1$
⇒ $h = \frac{1}{(\sqrt{3} - 1)}$
On multiplying the numerator and denominator of the above equation by $(\sqrt{3} + 1)$ , we get:
$h = \frac{1}{(\sqrt{3} - 1)} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)} = \frac{(\sqrt{3} + 1)}{3 - 1} = \frac{(\sqrt{3} + 1)}{2}$ $= \frac{1}{2} (\sqrt{3} + 1)$ km

Hence, the height of the hill is $\frac{1}{2} (\sqrt{3} + 1)$ km.

Page No 673:

Answer:

(c) $10 \sqrt{3} m$
Let $A B$ be the pole and $A C$ and $A D$ be its shadows.
We have:
∠ $A C B = 30^{o}$ , ∠ $A D B = 60^{o}$ and $A B = 15$ m

In ∆ $A C B$ , we have:
$\frac{A C}{A B} = c o t 30^{o} = \sqrt{3}$
⇒ $\frac{A C}{15} = \sqrt{3}$ ⇒ $A C = 15 \sqrt{3}$ m

Now, in ∆ $A D B$ , we have:
$\frac{A D}{A B} = c o t 60^{o} = \frac{1}{\sqrt{3}}$
⇒ $\frac{A D}{15} = \frac{1}{\sqrt{3}}$ ⇒ $A D = \frac{15}{\sqrt{3}}$ $= \frac{15 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{15 \sqrt{3}}{3} = 5 \sqrt{3}$ m

∴ Difference between the lengths of the shadows = $A C - A D = 15 \sqrt{3} - 5 \sqrt{3} = 10 \sqrt{3}$ m

Page No 674:

Answer:

(b) 30 m
Let $A B$ be the observer and $C D$ be the tower.

Draw $B E$ ⊥ $C D$ , Let $C D = h$ metres. Then,
$A B = 1.5$ m , $B E = A C = 28.5$ m and ∠ $E B D = 45^{o}$ .
$D E = (C D - E C)$ = $(C D - A B) = (h - 1.5)$ m.
In right ∆ $B E D$ , we have:
$\frac{D E}{B E} = \tan 45^{o} = 1$

⇒ $\frac{(h - 1.5)}{28.5} = 1$

⇒ $h - 1.5 = 28.5$
⇒ $h = 28.5 + 1.5 = 30$ m
Hence the height of the tower is $30$ m.

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