Rd Sharma 2022 for Class 10 Maths Chapter 12 - Heights And Distances

Answer:

Let be the tower of height m and C be the point on the ground, makes an angle of elevation with the top of tower.

In a triangle, given that BC = 20 m and

Now we have to find height of tower, so we use trigonometrical ratios.

In the triangle,

Hence height of tower is meters.

Answer:

Let be the ladder of length m and C be the points, makes an angle of elevation 60° with the wall and foot of the ladder is 9.5 meter away from wall.

In a triangle ABC, given that BC = 9.5 m and angle C = 60°

Now we have to find length of ladder.

So we use trigonometrically ratios.

In a triangle ABC,

Hence length of ladder is meters.

Answer:

Let the height of the wall be h meters.
In triangle ABC, 
$$\begin{gathered} \cos 60°=\frac{h}{15} \\ \Rightarrow \frac{1}{2}=\frac{h}{15} \\ \Rightarrow h=7.5 \mathrm{m} \end{gathered}$$
Hence, the height of the wall is 7.5 meters. 

Answer:

Let BC be the tower of height x m and AB be the flag staff of height y, 70 m away from the tower, makes an angle of elevation are 60° and 45° respectively from top and bottom of the flag staff.

Let AB = y m, BC = x m and CD = 70 m.

and

So we use trigonometric ratios.

In a triangle,

Again in a triangle,

Hence the height of flag staff is m and height of tower is m.

Answer:

Let CD be the tower. A and B are the two points on the same side of the tower.



$$\begin{gathered} \mathrm{In} ∆\mathrm{DBC}, \\ \tan 60°=\frac{\mathrm{DC}}{\mathrm{BC}} \\ \Rightarrow \sqrt{3}=\frac{15}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC}=\frac{15}{\sqrt{3}} \\ \Rightarrow \mathrm{BC}=5\sqrt{3} \mathrm{m} \end{gathered}$$
$$\begin{gathered} \mathrm{In} ∆\mathrm{DAC}, \\ \tan 45°=\frac{\mathrm{DC}}{\mathrm{AC}} \\ \Rightarrow 1=\frac{15}{\mathrm{AC}} \\ \Rightarrow \mathrm{AC}=15 \mathrm{m} \end{gathered}$$
Now,
AC = AB + BC
∴ AB = AC − BC$=15-5\sqrt{3}=5\left(3-\sqrt{3}\right) \mathrm{m}$ 
Hence, the distance between the two points A and B is $5\left(3-\sqrt{3}\right) \mathrm{m}$.

Answer:

Let BC be the tower of height h m and AB be the flag staff with distance 5m.Then angle of elevation from the top and bottom of flag staff are 60° and 30° respectively.

Let and,

Here we have to find height h of tower.

So we use trigonometric ratios.

In a triangle,

Again in a triangle

Hence the height of tree is m.

Answer:

Let h be height of tower and angle of elevation of foot of tower is 30°, on advancing
150 m towards the foot of tower then angle of elevation becomes 60°.

We assume that BC = x and CD = 150 m.

Now we have to prove height of tower is 129.9 m.

So we use trigonometrical ratios.

In a triangle,

Again in a triangle,

Hence the height of tower is m proved.

Answer:

Let h be height of tower and the angle of elevation as observed from the foot of tower is 32° and observed move towards the tower with distance 100 m then angle of elevation becomes 63°.

Let BC = x and CD = 100

Now we have to find height of tower

So we use trigonometrical ratios.

In a triangle ABC,

Again in a triangle,

So distance of the first position from the tower is = 100 + 46.7 = 146.7 m

Hence the height of tower is m and the desires distance is m.

Answer:

Let be height of tower and the angle of elevation of the top of tower from a point on the ground is and on moving with distance m towards the foot of tower on the pointis.

Let and

Now we have to find height of tower and distance of tower from point A.

So we use trigonometrical ratios.

In,

Again in ,

So distance

Hence the required height is m and distance is m.

Answer:

In the figure let OD = h and AD be the tower. The angle of elevation from the top of building to the top of tower is to be found 30°. Height of building ism and an angle of elevation from the bottom of same building is found to be 60°.

Let DC = x and , ,

Here we have to find height of tower and distance between the tower and building.

The corresponding diagram is as follows

In a triangle,

Again in a triangle,

So height of the tower is as follows:

Hence the required height is meter and distance is meter.

Answer:

Let AB be the tower of height h and AD be the flag pole on tower. At the point 9m away from the foot of tower, the angle of elevation of the top and bottom of flag pole are 60° and 30°. Let AD = x, BC = 9 and, .

Here we have to find height of tower and height of flag pole.

The corresponding diagram is as follows

In a triangle ABC,

Again in a triangle,

So height of tower is meter and height of flag pole is meters.

Answer:

Let be the observer of m tall. And be the tower of height. Here we have to find angle of elevation of the top of tower.

Let

The corresponding figure is as follows

In,

Hence the required angle is .

Answer:

Let BG be the distance of tall Boy x and he walks towards the building, makes an angle of elevation at top of building increase from 30° to 60°.

Therefore ∠A = 30° and ∠F = 60° given CE = 30 m, AB = 15 m, FG = 1.5 and DE = 28.5, GC = X − x and FD = X − x

We have to find x

The corresponding figure is as follows

In,

Again in,

Hence the required distance is m.

Answer:

Let be the tower of height.given the shadow of tower m. attitude of sun are and. Here we have to find height of tower. Let and .

So we have trigonometric ratios

In

Again in

Put

Hence height of tower is m.

Answer:

Let be the building of height m and the transmission tower of height meter.

Again let the angle of elevation of the bottom and top of tower at the point is 45° and 60° respectively.

In

Again in

Hence the height of tower is m.

Answer:

Let AD be the multistoried building of height h m. And angle of depression of the top and bottom are 30° and 45°. We assume that BE = 8, CD = 8 and BC = x, ED = x and
AC = h − 8. Here we have to find height and distance of building.

We use trignometrical ratio.

In,

Again in,

And

Hence the required height is meter and distance is meter.

Answer:

Let be the pedestal of height m and the statue of height meter and angle of elevation at top of statue is 60° and angle of elevation of pedestal at the same point is 45°. Here we have to find height of pedestal.

The corresponding figure is here

In ΔOAB

Again in ΔOAC

Hence the height of pedestal is .

Answer:

Let AB be the tower of height 120 m. 
C and D be two cars which are at angle of depression 45º and 60º respectively when observed from the top of the tower.
In $△$ADB,
$$\begin{gathered} \tan 60°=\frac{\mathrm{AB}}{\mathrm{DB}} \\ \Rightarrow \sqrt{3}=\frac{120}{\mathrm{DB}} \\ \Rightarrow \mathrm{DB}=\frac{120}{\sqrt{3}} .....\left(1\right) \end{gathered}$$
In $△$ACB,
$$\begin{gathered} \tan 45°=\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow 1=\frac{120}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC}=120 .....\left(2\right) \end{gathered}$$
Distance between the two cars will be
DB + BC = $\frac{120}{\sqrt{3}}+120$
$$\begin{gathered} \Rightarrow \mathrm{DB}+\mathrm{BC}=120\left(\frac{1}{\sqrt{3}}+1\right) \\ =120\left(\frac{1+\sqrt{3}}{\sqrt{3}}\right) \\ =120\times \frac{2.732}{1.732}=189.28 \mathrm{m} \end{gathered}$$

Answer:

Let OC be the tower of height H m and m high building makes an angle of elevation of top of cable wire is and an angle of depression from the its foot is .

Let, and, and,

So we use trigonometric ratios.

In a triangle,

Again in a triangle,

Hence the height of tower is m.

Answer:

Let be the height of light house m. and and the position of two ships and angle of depression are and. Let and,

Here we have to find distance between two ships.

The corresponding figure is as follows

So we trigonometric ratios,

In ΔOBC

Again in

Hence distance between two ships is m.

Answer:

Let AD be the building of height h m. and an angle of elevation of top of building from the foot of tower is 30° and an angle of the top of tower from the foot of building is 60°.

Let AD = h, AB = x and BC = 50 and ,

So we use trigonometric ratios.

In a triangle,

Again in a triangle,

Hence the height of building is m.

Answer:

Let be the width of river. And the angle of depression of the bank on opposite side of the river are 30° and 45° respectively. It is given thatm. Let and . And, .

Here we have to find the width of river.

We have the following figure

So we use trigonometric ratios.

In a triangle,

Again in a triangle,

So width of river is:

Hence the width of river is .

Answer:

Let be the height of hill. And be the tower of heightm. Angle of elevation of the top of hill from the foot of tower is 60° and angle of elevation of top of tower from foot of hill is 30°. Let and,

Here we have to find height of hill.

The corresponding figure is as follows

So we use trigonometric ratios.

In,

Again in

Hence the height of hill is m.

Answer:

Let be the height of light house. Angle of elevation of the top of light house from two boats are 30° and 45°. Let, and it is given thatm. So . And,

Here we have to find height of light house.

The corresponding figure is as follows

So we use trigonometric ratios.

In,

Again in

Hence the height of light house is m.

Answer:

Let AB and AD be the two men either side of cliff and height of cliff is 80 m.

And makes an angle of elevation, 30° and 60° respectively of the top of the cliff

We have given that AC = 80 m. Let BC = x and CD = y. And ,

Here we have to find height of cliff.

We have the corresponding figure as follows

So we use trigonometric ratios.

In a triangle,

Again in a triangle

Hence the height of cliff is m.

Answer:

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

Height of the aeroplane above the river, CD = 210 m

Now,

$\angle$CAD = $\angle$ADX = 60º         (Alternate angles)

$\angle$CBD = $\angle$BDY = 45º         (Alternate angles)

In right ∆ACD,

$$\begin{gathered} \tan 60°=\frac{\mathrm{CD}}{\mathrm{AC}} \\ \Rightarrow \sqrt{3}=\frac{210}{\mathrm{AC}} \\ \Rightarrow \mathrm{AC}=\frac{210}{\sqrt{3}}=70\sqrt{3} \mathrm{m} \end{gathered}$$

In right ∆BCD,

$$\begin{gathered} \tan 45°=\frac{\mathrm{CD}}{\mathrm{BC}} \\ \Rightarrow 1=\frac{210}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC}=210 \mathrm{m} \end{gathered}$$

∴ Width of the river, AB = BC + AC
                                   $$\begin{gathered} =210+70\sqrt{3} \\ =210+70\times 1.73 \\ =210+121.1 \\ =331.1 \mathrm{m} \end{gathered}$$

Hence, the width of the river is 331.1 m.

Answer:

Let be the tower height of m. flag height is m and an angle of elevation of top of tower is 45° and an angle of elevation of the top of flag is 60°.

Let, m and m and,

We have the corresponding angle as follows

So we use trigonometric ratios.

In a triangle,

Again in a triangle,

Hence the height of flag is m.

Answer:

Let AB be the first pole and CD be the second pole.

Distance between the two poles, BD = 15 m

Height of the second pole, CD = 24 m

Suppose the height of the first pole be h m.

Draw AE ⊥ CD.

∴ CE = CD − ED = (24 − h) m                       [AB = ED = h m]

AE = BD = 15 m

Now, $\angle$CAE = $\angle$ACF = 30º        (Alternate angles)

In right ∆ACE,

$$\begin{gathered} \tan 30°=\frac{\mathrm{CE}}{\mathrm{AE}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{24-h}{15} \\ \Rightarrow \frac{15}{\sqrt{3}}=24-h \\ \Rightarrow h=24-5\sqrt{3} \end{gathered}$$

$\Rightarrow h=24-5\times 1.732=15.34 \mathrm{m}$

Hence, the height of the first pole is 15.34 m.

Answer:

Let CD be the the light house and A and B be the positions of the two ships.

AB = 200 m    (Given)

Suppose CD = h m and BC = x m

Now,

$\angle$DAC = $\angle$ADE = 30º        (Alternate angles)

$\angle$DBC = $\angle$EDB = 45º        (Alternate angles)

In right ∆BCD,

$$\begin{gathered} \tan 45°=\frac{\mathrm{CD}}{\mathrm{BC}} \\ \Rightarrow 1=\frac{h}{x} \\ \Rightarrow x=h .....\left(1\right) \end{gathered}$$

In right ∆ACD,

$$\begin{gathered} \tan 30°=\frac{\mathrm{CD}}{\mathrm{AC}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+200} \\ \Rightarrow \sqrt{3}h=x+200 .....\left(2\right) \end{gathered}$$

From (1) and (2), we get

$$\begin{gathered} \sqrt{3}h=200+h \\ \Rightarrow \sqrt{3}h-h=200 \\ \Rightarrow \left(\sqrt{3}-1\right)h=200 \\ \Rightarrow h=\frac{200}{\sqrt{3}-1} \end{gathered}$$
$$\begin{gathered} \Rightarrow h=\frac{200\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)} \\ \Rightarrow h=\frac{200\left(\sqrt{3}+1\right)}{2}=100\left(\sqrt{3}+1\right) \mathrm{m} \end{gathered}$$

Hence, the height of the light house is $100\left(\sqrt{3}+1\right)$ m.

Answer:

Let be tower of height m and angle of elevation of the top of tower from two points are and

Let, m and m and

The corresponding figure is as follows

So we use trigonometric ratios.

In,

Again in,

Hence the height of tower is m.

Answer:

Let AB be the tower and CD be the chimney.

Height of the tower, AB = 40 m

Suppose the height of the chimney be h m.

Draw AE ⊥ CD.

Here, CE = AB = 40 m

DE = CD − CE = (h − 40) m

In right ∆AEC,

$$\begin{gathered} \tan 30°=\frac{\mathrm{CE}}{\mathrm{AE}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{40}{\mathrm{AE}} \\ \Rightarrow \mathrm{AE}=40\sqrt{3} \mathrm{m} \end{gathered}$$

In right ∆AED,

$$\begin{gathered} \tan 60°=\frac{\mathrm{DE}}{\mathrm{AE}} \\ \Rightarrow \sqrt{3}=\frac{h-40}{40\sqrt{3}} \\ \Rightarrow h-40=40\sqrt{3}\times \sqrt{3}=120 \\ \Rightarrow h=120+40=160 \mathrm{m} \end{gathered}$$

Thus, the height of the chimney is 160 m.

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

We should follow the pollution control norms and contribute to the cleaniness of the environment.

Answer:

Let be the building of height 60 and be the lamp post of height, an angle of depression of the top and bottom of vertical lamp post are 30° and 60° respectively. Let, and . It is also given m. Then And ,

We have to find the following

(i) The horizontal distance between AB and CD

(ii) The height of lamp post

(iii) The difference between the heights of building and the lamp post

We have the corresponding figure as follows

(i) So we use trigonometric ratios.

In

Hence the distance between and is

(ii) Again in

Hence the height of lamp post is m.

(iii) Since BE = 60 − h

Hence the difference between height of building and lamp post is m.

Answer:

​Let the distance BC be x m and CD be y m. 


$$\begin{gathered} \mathrm{In} ∆\mathrm{ABC}, \\ \tan 60°=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{150}{x} \\ \Rightarrow \sqrt{3}=\frac{150}{x} \\ \Rightarrow x=\frac{150}{\sqrt{3}} \mathrm{m} .....\left(1\right) \end{gathered}$$

$$\begin{gathered} \mathrm{In} ∆\mathrm{ABD}, \\ \tan 45°=\frac{\mathrm{AB}}{\mathrm{BD}}=\frac{150}{x+y} \\ \Rightarrow 1=\frac{150}{x+y} \\ \Rightarrow x+y=150 \\ \Rightarrow y=150-x \end{gathered}$$

Using (1), we get

$\Rightarrow y=150-\frac{150}{\sqrt{3}}=\frac{150\left(\sqrt{3}-1\right)}{\sqrt{3}}\mathrm{m}$
Time taken to move from point C to point D is 2 min = $\frac{2}{60} \mathrm{h}=\frac{1}{30} \mathrm{h}$.
Now,

$\mathrm{Speed}=\frac{\mathrm{Distance}}{\mathrm{Time}}=\frac{y}{{\displaystyle \frac{1}{30}}}$
$=\frac{\frac{150\left(\sqrt{3}-1\right)}{\sqrt{3}}}{{\displaystyle \frac{1}{30}}}=1500\sqrt{3}\left(\sqrt{3}-1\right) \mathrm{m}/\mathrm{h}$

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,

$$\begin{gathered} \tan 60°=\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow \sqrt{3}=\frac{100}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC}=\frac{100}{\sqrt{3}}=\frac{100\sqrt{3}}{3} \mathrm{m} \end{gathered}$$

In right ∆ABD,

$$\begin{gathered} \tan 30°=\frac{\mathrm{AB}}{\mathrm{BD}} \\ \Rightarrow \tan 30°=\frac{\mathrm{AB}}{\mathrm{CD}+\mathrm{BC}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{2v+{\displaystyle \frac{100\sqrt{3}}{3}}} \end{gathered}$$
$$\begin{gathered} \Rightarrow 2v+\frac{100\sqrt{3}}{3}=100\sqrt{3} \\ \Rightarrow 2v=100\sqrt{3}\times \left(1-\frac{1}{3}\right) \\ \Rightarrow 2v=100\sqrt{3}\times \frac{2}{3} \end{gathered}$$
$$\begin{gathered} \Rightarrow v=\frac{100\sqrt{3}}{3} \\ \Rightarrow v=\frac{100\times 1.732}{3}=57.73 \mathrm{m}/\min \end{gathered}$$

Hence, the speed of the boat is 57.73 m/min.

Answer:

Let C′ be the image of cloud C. We have and.

Again let .and be the distance of cloud from point of observation.

We have to prove that

The corresponding figure is as follows

We use trigonometric ratios.

In

Again in

Now,

Again in

Hence distance of cloud from points of observation is

Answer:

Let be the height of aero plane above the road. And and be the two consecutive milestone, thenmile. We have and.

We have to prove that

The corresponding figure is as follows

In

Again in

Now,

Hence height of aero plane is

Answer:

Let be the ladder such that its top is on the wall and bottom is on the ground. The ladder is pulled away from the wall through a distance a, so that its top slides and takes position. So

And. Let

We have to prove that

We have the corresponding figure as follows

We use trigonometric ratios.

In

And

Again in

And

Now,

And

So

Hence .

Answer:

Let be the height of light house. And an angle of depression of the top of light house from two ships are and respectively. Let, . And , .

We have to find distance between the ships

We have the corresponding figure as follows

We use trigonometric ratios.

In

Again in

Now,

Hence the distance between ships is .

Answer:

Let the two objects be at points C and D. 
The angle of depression for the point C is $\beta$ and for the point D is $\alpha$.
In ∆ABC,
$$\begin{gathered} \tan \beta =\frac{h}{BC} \\ \Rightarrow BC=\frac{h}{\tan \beta } \end{gathered}$$
In ∆ABD,
$$\begin{gathered} \tan \alpha =\frac{h}{BD} \\ \Rightarrow \tan \alpha =\frac{h}{BC+CD} \\ \Rightarrow \tan \alpha =\frac{h}{{\displaystyle \frac{h}{\tan \beta }}+CD} \\ \Rightarrow \frac{h}{\tan \beta }+CD=\frac{h}{\tan \alpha } \end{gathered}$$
$$\begin{gathered} \Rightarrow CD=\frac{h}{\tan \alpha }-\frac{h}{\tan \beta } \\ \Rightarrow CD=h\left[\frac{1}{\tan \alpha }-\frac{1}{\tan \beta }\right] \\ \Rightarrow CD=h\left[cot\alpha -cot\beta \right] \end{gathered}$$
 

Answer:

DISCLAIMER: The question given in the book is incorrect. We are proving the height of the house to be $h\left(1+\tan \alpha cot\beta \right)$

Let the window be at point A. 
To prove:  BD = $h\left(1+\tan \alpha cot\beta \right)$
In ∆ABC,
$$\begin{gathered} \tan \alpha =\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{BC}{x} \\ \Rightarrow x=BC\tan \alpha .....\left(\mathrm{i}\right) \end{gathered}$$
In ∆ACD,
$$\begin{gathered} \tan \beta =\frac{CD}{AC}=\frac{h}{x} \\ \Rightarrow x=\frac{h}{\tan \beta } .....\left(\mathrm{ii}\right) \end{gathered}$$
From (i) and (ii)
$$\begin{gathered} BC\tan \alpha =\frac{h}{\tan \beta } \\ \Rightarrow BC=\frac{h\tan \alpha }{\tan \beta } \end{gathered}$$
Height of house = BC + CD
$$\begin{gathered} =\frac{h\tan \alpha }{\tan \beta }+h \\ =h\left[\frac{\tan \alpha }{\tan \beta }+1\right] \\ =h\left[\tan \alpha cot\beta +1\right] \end{gathered}$$
 

Answer:

Let the window G be 2 m above the ground. 
Window A be 4 m above the window G.
Balloon be at point B above the ground.
In ∆ABC, 
$$\begin{gathered} \tan 30°=\frac{h}{x} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x} \\ \Rightarrow x=h\sqrt{3} .....\left(\mathrm{i}\right) \end{gathered}$$
In ∆BGD,
$$\begin{gathered} \tan 60°=\frac{BD}{GD} \\ \Rightarrow \sqrt{3}=\frac{h+4}{x} \\ \Rightarrow x=\frac{h+4}{\sqrt{3}} .....\left(\mathrm{ii}\right) \end{gathered}$$
From (i) and (ii) we get,
$$\begin{gathered} h\sqrt{3}=\frac{h+4}{\sqrt{3}} \\ \Rightarrow 3h=h+4 \\ \Rightarrow 2h=4 \\ \Rightarrow h=2 \end{gathered}$$
Hence, the height of the ballon above the ground = 2 + 4 + 2 = 8 m

Answer:

Let be the length of shadow is m

Given that: Height of tower is meters and altitude of sun is

Here we have to find length of shadow.

So we use trigonometric ratios.

In a triangle,

Hence the length of shadow is m.

Answer:

Let be the angle of elevation of sun is.

Given that: Height of tower is meters and length of shadow is 1.

Here we have to find angle of elevation of sun.

In a triangle,

Hence the angle of elevation of sun is .

Answer:

Let be the angle of elevation of sun is.

Given that: Height of pole is meters and length of shadow is meters. Because length of shadow is equal to the height of pole.

Here we have to find angle of elevation of sun.

So we use trigonometric ratios.

In a triangle,

Hence the angle of elevation of sun is .

Answer:

Let be the height of tower is meters.

Given that: angle of elevation is and meters.

Here we have to find height of tower.

So we use trigonometric ratios.

In a triangle,

Hence height of tower is .

Answer:

Let be the height of tower is meters.

Given that: angle of elevation are and and also m and m.

Here we have to find height of tower.

So we use trigonometric ratios.

In a triangle,

Again in a triangle ,

Put

Hence height of tower is meters.

Answer:

In a triangle,

We know that

Again,

In a triangle,

Hence the required angles are .

$\epsilon$

Answer:

Let AB and CD be the two towers and E be the mid-point of AC.

Height of the tower, AB = y

Height of the tower, CD = x

It is given that, $\angle \mathrm{AEB}=60°$ and $\angle \mathrm{CED}=30°$.

Also, AE = EC

In right ∆AEB,

$$\begin{gathered} \tan 60°=\frac{\mathrm{AB}}{\mathrm{AE}} \\ \Rightarrow \sqrt{3}=\frac{y}{\mathrm{AE}} \\ \Rightarrow \mathrm{AE}=\frac{y}{\sqrt{3}} \end{gathered}$$

In right ∆CED,

$$\begin{gathered} \tan 30°=\frac{\mathrm{CD}}{\mathrm{CE}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{x}{\mathrm{CE}} \\ \Rightarrow \mathrm{CE}=\sqrt{3}x \end{gathered}$$

Now, AE = CE

$$\begin{gathered} \frac{y}{\sqrt{3}}=\sqrt{3}x \\ \Rightarrow y=3x \\ \Rightarrow \frac{x}{y}=\frac{1}{3} \\ \therefore x:y=1:3 \end{gathered}$$

Hence, the ratio of x : y is 1 : 3.

Answer:

Let the height of the tower AB be h units.

Suppose C is a point on the ground such that $\angle \mathrm{ACB}=30°$.

In right ∆ACB,

$$\begin{gathered} \tan 30°=\frac{\mathrm{AB}}{\mathrm{AC}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{\mathrm{AC}} \\ \Rightarrow \mathrm{AC}=\sqrt{3}h .....\left(1\right) \end{gathered}$$

Let the angle of elevation of the top of the tower at C be θ, if the height of the tower is tripled.

New height of the tower, AD = 3h units

In right ∆ACD,

$$\begin{gathered} \tan \theta =\frac{\mathrm{AD}}{\mathrm{AC}} \\ \Rightarrow \tan \theta =\frac{3h}{\mathrm{AC}} \\ \Rightarrow \tan \theta =\frac{3h}{\sqrt{3}h}=\sqrt{3} \left[\mathrm{From} \left(1\right)\right] \\ \Rightarrow \tan \theta =\tan 60° \\ \Rightarrow \theta =60° \end{gathered}$$

Hence, the required angle of elevation is 60º.

Answer:

In the given figure,
AB = AD + DB = 6 m
Given: AD = 2.54 m
⇒ 2.54 m + DB = 6 m
⇒ DB = 3.46 m

Now, in the right triangle BCD,
$\frac{\text{BD}}{\text{CD}}=\sin 60°$
$$\begin{gathered} \Rightarrow \frac{3.46 \text{m}}{\text{CD}}=\frac{\sqrt{3}}{2} \\ \Rightarrow \frac{3.46 \text{m}}{\text{CD}}=\frac{1.73}{2} \\ \Rightarrow \text{CD}=\frac{2\times 3.46 \text{m}}{1.73} \\ \Rightarrow \text{CD}=4 \text{m} \end{gathered}$$

Thus, the length of the ladder CD is 4 m.

Answer:

Let AB be the height of the observer and EC be the height of the tower.

Given:
AB = 1.7 m ⇒ CD = 1.7 m
BC = 20$\sqrt{3}$ m

Let ED be h m.



In ∆ADE,

$$\begin{gathered} \tan 30°=\frac{\mathrm{ED}}{\mathrm{AD}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{20\sqrt{3}} \\ \Rightarrow h=20 \mathrm{m} \end{gathered}$$

∴ EC = ED + DC = (h + 1.7) m = 21.7 m

Hence, the height of the tower is 21.7 m.

Answer:

Let AB = 1.5 m be the observer and CD = 30 m be the tower.
Let the angle of elevation of the top of the tower be $\alpha$.
CD = CE + ED
$$\begin{gathered} \Rightarrow \mathrm{CD}=\mathrm{CE}+\mathrm{AB} \\ \Rightarrow 30=\mathrm{CE}+1.5 \\ \Rightarrow \mathrm{CE}=30-1.5=28.5 \mathrm{m} \end{gathered}$$
In $△$CEB,
$$\begin{gathered} \tan \alpha =\frac{CE}{BE}=\frac{28.5}{28.5} \\ \Rightarrow \tan \alpha =1 \\ \Rightarrow \tan \alpha =\tan 45° \\ \Rightarrow \alpha =45° \end{gathered}$$

Answer:

Given that $\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{1}{\sqrt{3}}$

From the figure, it is clear that $△$ABC is a right-angled triangle in which AB is the vertical rod and BC is its shadow.
We have,
$$\begin{gathered} \tan \theta =\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{1}{\sqrt{3}} \\ \Rightarrow \tan \theta =\tan 30° \\ \Rightarrow \theta =30° \end{gathered}$$
Hence, the required angle of elevation of the sun is 30$°$.

Answer:

Let the angle of elevation of Sun be θ.



Here, AB is the pole and BC is the shadow of the pole.

AB = 6 m and BC = $2\sqrt{3}$ m

In right ∆ABC,

$$\begin{gathered} \tan \theta =\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow \tan \theta =\frac{6 \mathrm{m}}{2\sqrt{3} \mathrm{m}} \\ \Rightarrow \tan \theta =\sqrt{3} \end{gathered}$$
$$\begin{gathered} \Rightarrow \tan \theta =\tan 60° \\ \Rightarrow \theta =60° \end{gathered}$$

Hence, the Sun's elevation is 60º.

If a pole 6 m high casts a shadow $2\sqrt{3}$ m long on the ground, then the Sun's elevation is ___60º___.

Answer:

Let the angle of elevation of Sun be θ.



Here, AB is the pole and BC is the shadow of the pole.

AB = h m and BC = $\sqrt{3}h$ m

In right ∆ABC,

$$\begin{gathered} \tan \theta =\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow \tan \theta =\frac{h \mathrm{m}}{\sqrt{3}h \mathrm{m}} \\ \Rightarrow \tan \theta =\frac{1}{\sqrt{3}} \end{gathered}$$
$$\begin{gathered} \Rightarrow \tan \theta =\tan 30° \\ \Rightarrow \theta =30° \end{gathered}$$

Hence, the angle of elevation of Sun is 30º.

The angle of elevation of the sun when the shadow of a pole h meter high is $\sqrt{3}h$ meters long is ____30º____.

Answer:

Let AB be the height of the tower and the initial distance of point of observation from its foot be BC.

Suppose the initial angle of elevation from the point of observation be θ.



In right ∆ABC,

$\tan \theta =\frac{\mathrm{AB}}{\mathrm{BC}} .....\left(1\right)$

If the height of a tower and the distance of the point of observation from its foot, both, are increased by 10%, then

New height of the tower, A'B' = AB + 10% of AB $=\mathrm{AB}+\frac{1}{10}\mathrm{AB}=\frac{11}{10}\mathrm{AB}$
New distance of point of observation from the foot, B'C' = BC + 10% of BC $=\mathrm{BC}+\frac{1}{10}\mathrm{BC}=\frac{11}{10}\mathrm{BC}$

Suppose the new angle of elevation from the point of observation be θ'.



In right ∆A'B'C',

$$\begin{gathered} \tan \theta \text{'}=\frac{\mathrm{A}\text{'}\mathrm{B}\text{'}}{\mathrm{B}\text{'}\mathrm{C}\text{'}} \\ \Rightarrow \tan \theta \text{'}=\frac{{\displaystyle \frac{11}{10}}\mathrm{AB}}{{\displaystyle \frac{11}{10}}\mathrm{BC}}=\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow \tan \theta \text{'}=\tan \theta \left[\mathrm{Using} \left(1\right)\right] \\ \Rightarrow \theta \text{'}=\theta \end{gathered}$$

So, the new angle of elevation of the top of the tower remains unchanged.

If the height of a tower and the distance of the point of observation from its foot, both, are increased by 10%, then the angle of elevation of its top remains __unchanged__.

Answer:

Let AB be the pole. Suppose BD and BC be the lengths of the shadow of the pole when the elevation of Sun is 30° and 60°, respectively.



Here, AB = 15 m

In right ∆ABC,

$$\begin{gathered} \tan 60°=\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow \sqrt{3}=\frac{15}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC}=\frac{15}{\sqrt{3}}=5\sqrt{3} \mathrm{m} .....\left(1\right) \end{gathered}$$

In right ∆ABD,

$$\begin{gathered} \tan 30°=\frac{\mathrm{AB}}{\mathrm{BD}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{15}{\mathrm{BD}} \\ \Rightarrow \mathrm{BD}=15\sqrt{3} \mathrm{m} .....\left(2\right) \end{gathered}$$

Now,

Difference between the length of the shadows = BD − BC $=15\sqrt{3}-5\sqrt{3}=10\sqrt{3} \mathrm{m}$

If the elevation of the sun changes from 30° to 60°, then the difference between the lengths of shadows of a pole 15 m high is $\underline{ 10\sqrt{3} \mathrm{m} }$.

Answer:

Let AB be the tower of height h m. Suppose C and D be the positions on the level ground such that CD = 20 m.



In right ∆ABC,

$$\begin{gathered} \tan 60°=\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow \sqrt{3}=\frac{h}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC}=\frac{h}{\sqrt{3}} \end{gathered}$$

In right ∆ABD,

$$\begin{gathered} \tan 30°=\frac{\mathrm{AB}}{\mathrm{BD}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{\mathrm{BD}} \\ \Rightarrow \mathrm{BD}=\sqrt{3}h \end{gathered}$$

Now,

BD − BC = 20 m

$$\begin{gathered} \Rightarrow \sqrt{3}h-\frac{h}{\sqrt{3}}=20 \\ \Rightarrow \frac{3h-h}{\sqrt{3}}=20 \\ \Rightarrow 2h=20\sqrt{3} \\ \Rightarrow h=10\sqrt{3} \mathrm{m} \end{gathered}$$

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

On the level ground, the angle of elevation of a tower is 30°. On moving 20 meters nearer, the angle of elevation is 60°. The height of the tower is $\underline{ 10\sqrt{3} \mathrm{m} }$.