Rs Aggarwal 2020 2021 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 2020 2021 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 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

#### Page No 656:

Let $AB$ be the tower standing vertically on the ground and O be the position of the observer.
We now have:
and ∠$AOB$
Let:
m Now, in the right ∆$OAB$, we have:
= $\sqrt{3}$

⇒

⇒  =

Hence, the height of the pole is 34.64 m.

#### Page No 657:

Let  be the horizontal ground and $A$ be the position of the kite.
Also, let O be the position of the observer and $OA$ be the thread.
Now, draw ⊥ $OX$.
We have:
${60}^{o}$m and ∠
Height of the kite from the ground = $AB$ = 75 m
Length of the string,  m In the right ∆$OBA$, we have:

⇒  m
Hence, the length of the string is $86.6$ m.

#### Page No 657: Let CE and AD be the heights of the observer and the chimney, respectively.

We have,

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

#### Page No 657: Let the height of the tower be AB.

We have,

So, the height of the tower is 10 m.

#### Page No 657: Let BC and CD be the heights of the tower and the flagstaff, respectively.

We have,

So, the height of the flagstaff is 87.8 m.

#### Page No 657: Let BC be the tower and CD be the water tank.

#### Page No 657: Let AB be the tower and BC be the flagstaff.

We have,

So, the height of the tower is 3 m.

#### Page No 657:

Let $AC$ be the pedestal and $BC$ be the statue such that  m.
We have:
and ∠
Let:
m and  m In the right ∆$ADC$, we have:

⇒
⇒
Or,

Now, in the right ∆$ADB$, we have:

⇒

On putting  in the above equation, we get:

⇒
⇒
⇒ m

Hence, the height of the pedestal is 2 m.

#### Page No 657:

Let $AB$ be the unfinished tower, $AC$ be the raised tower and O be the point of observation.
We have:
m, ∠ and ∠
Let  m such that  m. In ∆AOB, we have:

⇒  m =  m
In ∆$AOC$, we have:
$=\sqrt{3}$

⇒
m

∴ Required height =  m = 86.6 m

#### Page No 657:

Let $OX$ be the horizontal plane, be the tower and $CD$ be the vertical flagpole.
We have:
m, ∠ and ∠
Let:
m and  m In the right ∆$ABD$, we have:

⇒ $\frac{h}{9}=\frac{1}{\sqrt{3}}$
⇒  m
Now, in the right ∆$ABC$, we have:

⇒
⇒

By putting  in the above equation, we get:

⇒
⇒

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

#### Page No 657: Let AB and CD be the equal poles; and BD be the width of the road.

We have,

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:

Let $CD$ be the tower and  be the positions of the two men standing on the opposite sides. Thus, we have:
, ∠ and  m
Let  m and  m such that  m. In the right ∆$DBC$, we have:

⇒
⇒  m

In the right ∆$ACD$, we have:

⇒
⇒
On putting  in the above equation, we get:

⇒  m

∴ Distance between the two men =  m

#### Page No 658: Let PQ be the tower.

We have,

So, the distance between the cars is 273 m.

#### Page No 658: Let PQ be the tower.

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

#### Page No 658: Let PQ=h m be the height of the TV tower and BQ=x m be the width of the canal.

We have,

So, the height of the TV tower is  and the width of the canal is 10 m.

#### Page No 658: Let AB be the building and PQ be the tower.

We have,

So, the height of the building is 20 m.

#### Page No 658:

Let $DE$ be the first tower and $AB$ be the second tower.
Now,  m and  m such that  m and ∠.
Let  m such that  m and  m. In the right ∆$BCE$, we have:

⇒
⇒
⇒
⇒  =  m
∴ Height of the first tower =  m

#### Page No 658: Let PQ be the chimney and AB be the tower.

We have,

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

We have,

So, the height of the tower is 19.12 m.

#### Page No 659: Let PQ be the tower.

We have,

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

#### Page No 659: Let PQ be the tower.

We have,

So, the height of the tower is 15 m.

#### Page No 659: Let AD be the tower and BC be the cliff.

We have,

So, the height of the tower is 43.92 m.

#### Page No 659:

Let $AB$ be the deck of the ship above the water level and $DE$ be the cliff.
Now,
m such that  m and ∠​ and ∠.
If AD = x m and  m, then  m. In the right ∆$BAD$, we have:

⇒

⇒   m

In the right ∆$EBC$, we have:

⇒
⇒
⇒         [∵ ]
m

∴ Distance of the cliff from the deck of the ship =  m
And,
Height of the cliff =  m

#### Page No 659: We have,

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

#### Page No 659: Let the height of flying of the aeroplane be PQ = BC and point A be the point of observation.

We have,

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

#### Page No 659:

Let $AB$ be the tower.
We have:
m, ∠ and ∠
Let:
m  and  m In the right ∆$ABD$, we have:

⇒
⇒
Now, in the right ∆$ACB$, we have:

⇒
⇒

On putting  in the above equation, we get:

⇒
⇒
⇒  m
Hence, the height of the tower is 129.9 m.

#### Page No 659:

Let $OA$ be the lighthouse and B and C be the two positions of the ship.
Thus, we have:
m, ∠ and  ∠ Let:
m and  m
In the right ∆$OAC$, we have:

⇒ m
Now, in the right ∆$OBA$, we have:

⇒

On putting  in the above equation, we get:
m

∴ Distance travelled by the ship during the period of observation =  m

#### Page No 659: 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:
In the right ∆$APD$, we have:

⇒
⇒  m
In the right ∆$PDB$, we have:

⇒
⇒  m

∴ Width of the river =  m

#### Page No 659:

Let $AB$ be the tower and  be two points such that  m  and  m.
Let:
m, ∠ and ∠ In the right ∆BCA, we have:

In the right ∆BDA, we have:

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

⇒ 36 = h2
h = ±6

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

#### Page No 660: 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,

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

#### Page No 660: Let OP be the tower and points A and B be the positions of the cars.

We have,

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: Let AC be the pole and BD be the ladder.

We have,

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

#### Page No 660: We have,

#### Page No 660:

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.

Substituting the value of h from equation (i) in equation (ii), we get

#### Page No 660:

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,
$\angle$CAD = $\angle$ADX = 60º       (Alternate angles)
$\angle$CBD = $\angle$BDY = 45º        (Alternate angles)
In right ∆ACD,

In right ∆BCD,

∴ Width of the river, AB
= BC + AC

Thus, the width of the river is 473 m.

#### Page No 660: 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

Also, in

#### Page No 660:

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. From equation (i) and (ii) we get,

Hence, the height of the hill is 1.365 km.

#### Page No 661: 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 $\sqrt{3}:1.$
$⇒\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\sqrt{3}}{1}.$
In

#### Page No 661: 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,

In right ∆ABD,

$⇒2v+\frac{100\sqrt{3}}{3}=100\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒2v=100\sqrt{3}-\frac{100\sqrt{3}}{3}\phantom{\rule{0ex}{0ex}}⇒2v=100\sqrt{3}\left(1-\frac{1}{3}\right)\phantom{\rule{0ex}{0ex}}⇒2v=100\sqrt{3}×\frac{2}{3}$

Thus, the speed boat is 57.73 m/min.

#### Page No 661:

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,

In right ∆AED,

$⇒\sqrt{3}h-h=8\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒\left(\sqrt{3}-1\right)h=8\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒h=\frac{8\sqrt{3}}{\sqrt{3}-1}\phantom{\rule{0ex}{0ex}}⇒h=\frac{8\sqrt{3}\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)×\left(\sqrt{3}+1\right)}$

∴ Height of the tower =

Also,

Distance between the tower and building = BC =                  (BC = h m)

Thus, the the height of the tower is and the distance between the tower and the building is .

#### Page No 671: Let AB represents the vertical pole and BC represents the shadow on the ground and θ represents angle of elevation the sun.

Hence, the correct answer is option (c).

#### Page No 671: Here, AO be the pole; BO be its shadow and $\theta$ be the angle of elevation of the sun.

Hence, the correct answer is option (c).

#### Page No 671:

(b) 30°

Let $AB$ be the pole and $BC$ be its shadow. Let  and   such that  (given) and $\theta$ be the angle of elevation.
From ∆$ABC$, we have:

⇒
⇒
⇒

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

#### Page No 671: Let AB be the pole, BC be its shadow and $\theta$ be the sun's elevation.

Hence, the correct answer is option (a).

#### Page No 671: Let AB be a stick and BC be its shadow; and PQ be the tree and QR be its shadow.

We have,

Hence, the correct answer is option (d).

#### Page No 672: Let AB be the wall and AC be the ladder.

We have,

Hence, the correct answer is option (d).

#### Page No 672: Let AB be the wall and AC be the ladder.

We have,

Hence, the correct answer is option (c).

#### Page No 672: Let AB be the tower and point C be the point of observation on the ground.

We have,

Hence, the correct answer is option (b).

#### Page No 672: Let AB be the tower and point C be the position of the car.

We have,

Hence, the correct answer is option (b).

#### Page No 672: Let point A be the position of the kite and AC be its string.

We have,

Hence, the correct answer is option (b).

#### Page No 672: Let AB be the cliff and CD be the tower.

We have,

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

We have,

Hence, the correct answer is option (c).

#### Page No 672: Let CD = h be the height of the tower.

We have,

Hence, the correct answer is option (d).

#### Page No 672: Let AB be the rod and BC be its shadow; and $\theta$ be the angle of elevation of the sun.

Hence, the correct answer is option (a).

#### Page No 673: Let AB be the pole and BC be its shadow.

We have,

Hence, the correct answer is option (b).

#### Page No 673: Let the sun's altitude be $\theta$.

We have,

Hence, the correct answer is option (a).

#### Page No 673: Let AB and CD be the two towers such that AB = and CD = y.

We have,

Hence, the correct answer is option (c).

#### Page No 673:

(b) $10\sqrt{3}\mathrm{m}$
Let $AB$ be the tower and $O$ be the point of observation.
Also,
and  m
Let:
m In ∆$AOB$, we have:

m

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

#### Page No 673:

(a) $50\sqrt{3}\mathrm{m}$
Let $AB$ be the string of the kite and $AX$ be the horizontal line.
If $BC$ ⊥ $AX$, then  m and ∠.
Let:
m In the right ∆$ACB$, we have:

m
Hence, the height of the kite is $50\sqrt{3}$ m.

#### Page No 673:

(b) $\sqrt{ab}$
Let $AB$ be the tower and and $D$ be the points of observation on $AC$.
Let:
, ∠ and  m
Thus, we have:
and Now, in the right ∆ABC, we have:
⇒              ...(i)
In the right ∆ABD, we have:
⇒     ...(ii)

On multiplying (i) and (ii), we have:

[ ∵ ]
⇒
⇒  m

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

#### Page No 673:

(b) $10\sqrt{3}\mathrm{m}$
Let $AB$ be the tower and $C$ and $D$ be the points of observation such that ∠, ∠,  m  and  m. Now, in ∆$ADB$, we have:

⇒

In ∆$ACB$, we have:

⇒
∴
⇒
⇒  ⇒
∴ Height of the tower AB =  m

#### Page No 673:

(c) $16\sqrt{3}{\mathrm{cm}}^{2}$
Let $ABCD$ be the rectangle in which ∠ and  cm. In ∆$BAC$, we have:

⇒  m
Again,

m
∴ Area of the rectangle =  cm2

#### Page No 673:

(b) $\frac{1}{2}\left(\sqrt{3}+1\right)\mathrm{km}$
Let $AB$ be the hill making angles of depression at points $C$ and $D$ such that ∠, ∠ and  km.
Let:
km and  km In ∆$ADB$, we have:

⇒   ⇒         ...(i)
In ∆$ACB$, we have:

...(ii)
On putting the value of $h$ taken from (i) in (ii), we get:

On multiplying the numerator and denominator of the above equation by $\left(\sqrt{3}+1\right)$, we get:
km

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

#### Page No 673:

(c) $10\sqrt{3}\mathrm{m}$
Let $AB$ be the pole and $AC$ and $AD$ be its shadows.
We have:
, ∠ and m In ∆$ACB$, we have:

⇒   ⇒  m

Now, in ∆$ADB$, we have:

⇒   ⇒    m

∴ Difference between the lengths of the shadows =  m

#### Page No 674:

(b) 30 m
Let $AB$ be the observer and $CD$ be the tower. Draw $BE$ ⊥ $CD$, Let  metres. Then,
m ,  m and ∠.
=  m.
In right  ∆$BED$, we have:

⇒  m
Hence the height of the tower is $30$ m.

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