NCERT Solutions for Class 10 Maths Chapter 9 - Applications of Trigonometry
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Important Points
1. Angle of Elevation : The angle between the horizontal line and line of sight, is called angle of elevation.
2. Angle of depression : An angle that is formed with the horizontal line if the line of sight is downward from the horizontal line.
Exercise 9.1
1. A circus artist is clumbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° (see figure).
Sol. Given in the figure
AB (Height of the pole) = ?
AC (Length of Rope) = 20 m
In ΔABC,
$$\text{sin 30}\degree=\frac{\text{AB}}{\text{AC}}=\frac{\text{AB}}{\text{20}}\\\Rarr\space\frac{1}{2}=\frac{\text{AB}}{20}\\\Rarr\space\text{AB}=\frac{20}{2}=10 \text{m}$$
Hence, the height of the pole be 10 m.
2. A tree breaks due to storm and the broken part bends so that the top of the tree touches the gound making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find height of the tree.
Sol. Let the height of the tree be AC. When the storm cone, the tree broke from point B. The broken part of the tree BC touches the ground at point D, making an angle 30° on the ground.
Also, given AD = 8 m
In right ΔABD,
$$\text{tan 30\degree}=\frac{\text{AB}}{\text{AD}}\\\Rarr\space\frac{1}{\sqrt{3}}=\frac{\text{AB}}{8}\\\Rarr\space\text{AB}=\frac{8}{\sqrt{3}}\text{m}$$
Again in ΔAABO,
$$\text{cos 30\degree}=\frac{\text{AD}}{\text{BD}}\\\Rarr\frac{\sqrt{3}}{2}\\=\frac{8}{\text{BD}}\\\Rarr\space\text{BD}=\frac{16}{\sqrt{3}}$$
∴ AC = AB + BC = AB + BD
( ∵ BC = BD)
$$=\frac{8}{\sqrt{3}}+\frac{16}{\sqrt{3}}\\=\frac{24}{\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}}\\\frac{24\sqrt{3}}{3}=8\sqrt{3}\text{m}\\\text{Hence, the height of the tree is 8}\sqrt{3}\text{m.}$$
3. A contractor plans to install two slides for the children to play in a park, for the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
Sol. (i) In Fig. (a), it is the slide for the children below the age of 5 years. Let BC = 1.5 m be the height of the slides and slide AC is inclined at ∠CAB = 30° to the ground.
In right angled ΔABC,
$$\text{sin 30}\degree=\frac{\text{BC}}{\text{AC}}\\\Rarr\space\frac{1}{2}=\frac{1.5}{\text{AC}}\\\Rarr\text{AC}=3\text{m}$$
(ii) In Fig. (b), it is the slide for the elder children. Let RQ = 3 m be the height of the slides and slides PR is inclined at an ∠RPQ = 60° to the ground.
In right angled ΔQPR,
$$\text{sin 60}\degree=\frac{\text{PQ}}{\text{PR}}\\\Rarr\space\frac{\sqrt{3}}{2}=\frac{3}{\text{PR}}\\\Rarr\space\text{PR}=\frac{3×2}{\sqrt{3}}=2\sqrt{3}m$$
(ii) In Fig. (b), it is the slide for the elder children. Let RQ = 3 m be the height of the slides and slides PR is inclined at an ∠RPQ = 60° to the ground.
In right angled ΔQPR,
$$\text{sin 60}\degree=\frac{\text{RQ}}{\text{PR}}\\\Rarr\space\frac{\sqrt{3}}{2}=\frac{3}{\text{PR}}\\\Rarr\space\text{PR}=\frac{3×2}{\sqrt{3}}=2\sqrt{3}\text{m}\\\text{Hence, length of the slides}\\\text{ in each case are 3 m and}2\sqrt{3} m.$$
4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower is 30°. Find the height of the tower.
Sol. Let be in figure height of the tower BC = h
Distance between point and Tower AB = 30 m (Given)
Then, AB = 30 m
In right angled ΔABC,
$$\text{tan 30}\degree=\frac{\text{BC}}{\text{AB}}\\\Rarr\space\frac{1}{\sqrt{3}}=\frac{h}{30}\\\Rarr\space h=\frac{30}{\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}}\\=10\sqrt{3}\\\text{Hence, height of the tower is}\space 10\sqrt{3}\text{m.}$$
5. A kite ls flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.
Sol. Let be AC = x (Length of the string)
BC = 60 cm (Height of the kite on the ground)
∠BAC = 60°
In right angled ΔABC,
$$\text{sin 60}\degree=\frac{\text{BC}}{\text{AC}}\\\Rarr\space\frac{\sqrt{3}}{2}=\frac{60}{x}\\\Rarr\space x=\frac{60×2}{\sqrt{3}}\\\Rarr\space x=\frac{60×2}{\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}}\\x=\frac{120\sqrt{3}}{3}=40\sqrt{3}\text{m}\\\text{Hence, length of the string is}\\40\sqrt{3}\text{m.}$$
6. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as the walks towards the building. Find the distance he walked towards the building.
Sol. Let,
AB = Height of building
AB = 30 m
DC = Height of boy
DC = 1.5 m
In figure, DC = BF = 1.5
Then, AF = AB – BF
AF = 30 – 1.5 = 28.5 m
Let, DE = Waling distance of boy = x = ?
EF = y
Given, ∠ADF = 30°, ∠AEF = 60°
In right angled ΔADF,
$$\text{tan 30}\degree=\frac{\text{AF}}{\text{DF}}\\\text{but AF = 28.5 m,}\\ \text{DF = DE + EF = x + y}\\\Rarr\space\frac{1}{\sqrt{3}}=\frac{28.5}{x+y}\\\Rarr\space x+y=28.5\sqrt{3}\\...(i)$$
In right angled ΔBEF,
$$\text{tan 60\degree}=\frac{\text{AF}}{\text{EF}}\\\Rarr\space\sqrt{3}=\frac{28.5}{y}\\\Rarr\space y=\frac{28.5}{\sqrt{3}}\\\text{Put the value y in eq. (i),}\\x+\frac{28.5}{\sqrt{3}}=28.5\sqrt{3}\\\Rarr\space x=28.5\sqrt{3}=\frac{28.5}{\sqrt{3}}$$
$$\Rarr\space x=28.5\bigg(\sqrt{3}-\frac{1}{\sqrt{3}}\bigg)\\\Rarr\space x=28.5\bigg(\sqrt{3}-\frac{1}{\sqrt{3}}\bigg)\\\Rarr\space x=28.5\bigg(\frac{3-1}{\sqrt{3}}\bigg)\\\Rarr\space x=\frac{57}{\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}}\\\Rarr\space x=\frac{57\sqrt{3}}{3}=19\sqrt{3}\text{m.}\\\text{Hence, the waking distance is}\space 19\sqrt{3}\text{m.}$$
7. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
Sol. Let BC = 20 m (the height of the building) and DC = h metre (height of the tower, which is standing on the building). A be a fixed point on the ground.
Given also,
∠BAC = 45° and ∠BAD = 60°
Also, let AB = x m
In right angled ΔABC,
$$\text{tan 45\degree}=\frac{\text{BC}}{\text{AB}}\\\Rarr\space 1=\frac{20}{x}\qquad...\text{(i)}\\\Rarr\space x = 20 m\\\text{Again, in right angled} Δ\text{ABD,}\\\text{tan 60\degree}=\frac{\text{BD}}{\text{AB}}\\=\frac{\text{BC + CD}}{\text{AB}}\\\Rarr\space\sqrt{3}=\frac{20+h}{x}$$
Put the value x from eq. (i),
$$\Rarr\space\sqrt{3}=\frac{20+h}{20}\\\Rarr\space 20 + h = 20\sqrt{3}\\\Rarr\space h=20(\sqrt{3}-1)\text{m}\\\text{Hence, the height of the tower is}\\20(\sqrt{3}-1)\text{m.}$$
8. A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.
Sol. Let BC = h metre be the height of the pedestal and CD = 1.6 m be the length of the statue, which is standing on the pedestal. Point A be a fixed point on the ground. From the point A, the angle of elevations of the top of the statue and bottom of the statue are
∠DAB = 60° and ∠CAB = 45°
Also, let AB = x metre
In right angled ΔBAD,
$$\text{tan 60\degree}=\frac{\text{BD}}{\text{AB}}\\\Rarr\space\sqrt{3}=\frac{\text{CD+CB}}{x}$$
$$\text{tan 60\degree}=\frac{\text{BD}}{\text{AB}}\\\Rarr\space\sqrt{3}=\frac{\text{CD+CB}}{x}\\\Rarr\space h=\sqrt{3}x-1.6\space\text{...(i)}\\\text{In right ΔCAB,}\\\text{tan} 45\degree=\frac{\text{BC}}{\text{AB}}\\\Rarr\space 1=\frac{h}{x}$$
⇒ x = h
Putting x = h in eq. (i), we get
$$h=\sqrt{3}h-16\\\Rarr\space 4(\sqrt{3}-1)=16\\\Rarr\space h=\frac{16}{(\sqrt{3}-1)}×\frac{\sqrt{3}+1}{\sqrt{3}+1}\\\text{[From (a – b)(a + b) =} a^{2}-b^{2}]\\=\frac{16}{2}(\sqrt{3}+1)\\=8(\sqrt{3}+1)\text{m}\\\text{Hence, the length of the pedestal is}\\8(\sqrt{3}+1)\text{m}$$
9. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Sol. Let DC = 50 m be the height of the tower and AB = h metre be the height of the building, angle of elevation, from the bottom of building and tower as well are
∠DBC = 60° and ∠ACB = 30°
Also, let BC = x be the distance between foot of the tower and building.
In right angled ΔACB,
$$\text{tan 30}\degree=\frac{\text{AB}}{\text{BC}}\\=\frac{1}{\sqrt{3}}=\frac{h}{\sqrt{x}}\\\Rarr\space h=\frac{x}{\sqrt{3}}\space...\text{(i)}$$
Again, in right angled ΔDBC,
$$\text{tan 60\degree}=\frac{\text{DC}}{\text{DB}}\\\Rarr\space\sqrt{3}=\frac{50}{x}\\\Rarr\space x=\frac{50}{\sqrt{3}}\text{m}\\\text{Putting x = }\frac{50}{\sqrt{3}}\text{in eq. (i), we get}\\\text{h}=\frac{50}{\sqrt{3}}×\frac{1}{\sqrt{3}}\\=\frac{50}{3}=16\frac{2}{3}\text{m.}\\\text{Hence, the height of the building is}\\16\frac{2}{3}\text{m.}$$
10. Two poles of equal heights are standing opposite each other on either side of the road, which 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.
Sol. Let BD = 80 m be the width of the road. On both sides of the road poles AB = CD = h metre are standing. Let C be any point on BD such that point C makes an elevations are ∠ECD = 60° and ∠ACE = 30°.
Let DC = x, then BC = BD – CD = (80 – x) m
In right angled ΔACE,
$$\text{tan 30° =}\frac{\text{AB}}{\text{AC}}\\\Rarr\space\frac{1}{\sqrt{3}}=\frac{h}{80-x}\\\Rarr\space80 – x = h\sqrt3\\\Rarr h\sqrt{3}+x=80$$
Again, in right angled ΔECD.
$$\text{tan 60\degree}=\frac{\text{ED}}{\text{CD}}\\\Rarr\space\sqrt{3}=\frac{h}{x}\Rarr h=h\sqrt{3}x\space\text{...(ii)}\\\text{Putting h =}\sqrt{3}\text{xin eq.(i), we get}\\\sqrt{3}x×(\sqrt{3})+x=80$$
⇒ 3x + x = 80
⇒ 4x = 80
⇒ x = 20 m
Putting x = 20 m in eq. (ii), we get
$$h=20\sqrt{3}\text{m}$$
Also, BC = 80 – x
= 80 – 20 = 60 m
$$\text{Hence, height of the poles be 20}\sqrt{3}$$ and the distances of the point from the poles are 60 m and 20 m.
11. A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see figure). Find the height of the tower and the width of the canal.
Sol. Width of canal = BC = x (let)
Height of the tower = AB = h (let)
DC = 20 m (given)
In Δ ACB,
$$\text{tan 60° =}\frac{\text{AB}}{\text{AC}}\\\Rarr\space\sqrt{3}=\frac{h}{x}\\\Rarr\space\sqrt{3}x=h\space\text{...(i)}$$
In ΔADB,
$$\text{tan 30\degree}=\frac{\text{AB}}{\text{BD}}\\\Rarr\space\frac{1}{\sqrt{3}}=\frac{h}{\text{BC+CD}}\\\Rarr\space\frac{1}{\sqrt{3}}=\frac{h}{x+20}$$
Put the value h from eq. (i),
$$\frac{1}{\sqrt{3}}=\frac{\sqrt{3}x}{x+20}$$
⇒ x + 20 = 3x
⇒ 20 = 3x – x
⇒ 2x = 10 m
Put the value x in equation (i),
$$h=10\sqrt{3}\text{m}\\\text{Hence, the height of the tower is}\\10\sqrt{3}m\space\text{and width of the canal is 10 m.}$$
In figure,
AD = DC = 7 m
EC = Height of the tower
EC = h m (Let)
BC = distance between building and tower
BC = x = AD (Let)
∠CAD = ∠ACB = 45°
∠EAD = 60°
In ΔACB,
$$\text{tan 45}\degree=\frac{\text{AB}}{\text{AC}}\\\Rarr\space 1=\frac{7}{x}$$
⇒ x = 7 m
In ΔEAD,
$$\text{tan 60}\degree=\frac{\text{ED}}{\text{AD}}\\\Rarr\space \sqrt{3}=\frac{h-7}{x}\\\Rarr\space\sqrt{3}x=h-7\\\text{Put the value x.}\\\Rarr\space\sqrt{3}×7=h-7\\\Rarr\space h=7+7\sqrt{3}\\\Rarr\space h=7(1+\sqrt{3})\text{m}\\\text{Hence, the height of the tower is}\\7(1+\sqrt{3})\text{m.}$$
13. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Sol. Let AB = 75 m be the height of the lighthoue from the sea level. Let D and C be the position of two ships on the sea-level.
From point A of a lighthouse the angle of depression of two ships D and C are
∠OAD = 30° and ∠OAC = 45°
⇒ ∠ADB = 30° and ∠ACB = 45°
(Alternate angle)
Let distance between two ships DC = y metre and BC = x metre
In right angled ΔADB,
$$\text{tan 30}\degree=\frac{\text{AB}}{\text{CD}}\\\Rarr\space\frac{1}{\sqrt{3}}=\frac{75}{y+x}\\\Rarr\space\text{x + y = 75}\sqrt{3}\space\text{...(i)}$$
In right angled ΔACB,
$$\text{tan 45}\degree=\frac{75}{x}\\\Rarr\space 1=\frac{75}{x}$$
⇒ x = 75 m
Putting x = 75 m in eq. (i), we get
$$75+y=75\sqrt{3}\\\Rarr\space y=75(\sqrt{3}-1)\text{m}\\\text{Hence, the distance between two ships is}\\75(\sqrt{3}-1)\text{m.}$$
14. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 meter from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see figure). Find the distance travelled by the balloon during the interval.
Sol. Let AD = 12 m be the tall girl standing on the horizontal line AB. Let FH = EB = 88.2 m be the height of balloon the line AB. At the eye of the girl D, the angle of elevations are ∠FDC = 60° and ∠EDC = 30°.
Now, FG = EC = 88.2 – 12 = 87 m
Let the distance travelled bythe balloons HB = y metre and AH = x metre
∴ DG = x metre and GC = y metre
In right angled ΔFDG,
$$\tan 60\degree=\frac{\text{FG}}{\text{DG}}\\\Rarr\space\sqrt{3}=\frac{87}{x}\\\Rarr\space x=\frac{87}{\sqrt{3}}\space\text{...(i)}$$
In right angled ΔEDC,
$$\text{tan 30}\degree=\frac{\text{EC}}{\text{DC}}\\\Rarr\space\frac{1}{\sqrt{3}}=\frac{87}{\text{DG+GC}}\\\Rarr\space x+y=87\sqrt{3}\space...(\text{ii})\\\therefore\text{From eq. (i), putting} \\x=\frac{87}{\sqrt{3}}\text{in eq. (ii), we get}\\\frac{87}{\sqrt{3}}+y=87\sqrt{3}\\\Rarr\space y=87\bigg(\sqrt{3}-\frac{1}{\sqrt{3}}\bigg)$$
$$=\frac{87(3-1)}{\sqrt{3}}\\=\frac{87×2}{\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}}\\=\frac{87×2\sqrt{3}}{3}\\=29×2\sqrt{3}\\=58\sqrt{3}\space\text{m}\\\text{Hence, the distance between two balloons are}\\58\sqrt{3}\text{m}$$
15. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
Sol. Let CD = h metre be the height of the tower. At point D a man is standing on the tower and observe that a car at an angle of depression of 30°. After six second the angle of depression of the car is 60°.
i.e., ∠ODA = 30º and ∠ODB = 60°
⇒ ∠DAC = 30° and ∠DBC = 60°
(Alternate angle)
Let AB = y and BC = x
In right angled ΔBCD,
$$\text{tan 60\degree}=\frac{\text{CD}}{\text{BC}}\\\Rarr\space\sqrt{3}=\frac{h}{x}\\\Rarr\space h=\sqrt{3}\space x\\\text{In right angled ΔACD,}\\\text{tan 30\degree}=\frac{\text{CD}}{\text{AC}}\\\Rarr\space\frac{1}{\sqrt{3}}=\frac{h}{x+y}\\\Rarr\space x + y = h\sqrt{3}\\\Rarr\space x + y =\sqrt{3}x(\sqrt{3})$$
⇒ x + y = 3x
$$y=2x, x=\frac{y}{2}\text{m}$$
According to question
$$\text{Speed}=\frac{y}{6}\text{m/s}$$
Time taken by the car to reach the foot of the tower
$$\therefore\space\text{Time}=\frac{\text{Distance}}{\text{Speed}}\\\text{Time}=\frac{x×6}{y}\\\text{Put the value of x,}\\\text{T}=\frac{6y}{y×2}$$
T = 3 second
Hence, the car moves from point B to C in 3 s.
16. The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
Sol. Let CD = h metre be the height of the tower. AC be a horizontal line on a ground. A and B be the two points on a line at a distance of 9 m and 4 m from the base of the tower.
Let, ∠CBD = θ, then ∠CAD = 90° – θ
(The complementary means the sum of two angles are 90°)
In right angled ΔCAD,
$$\text{ṭan(90\degree}-\theta)=\frac{\text{CD}}{\text{AC}}\\\Rarr\space\text{cot}\theta=\frac{h}{9}\space\text{...(i)}$$
And in right angled ΔCBD,
$$\text{tan}\theta=\frac{\text{CD}}{\text{BC}}\\\Rarr\space\text{tan}\theta=\frac{h}{4}\\\frac{1}{\text{cot}\theta}=\frac{h}{4}\\\text{cot}\theta=\frac{4}{h}\space\text{..(ii)}$$
By equation (i) and (ii),
$$\frac{h}{9}=\frac{4}{h}$$
⇒ h2 = 36
⇒ h = 6 m
Hence, the height of the tower is 6 m.
Hence proved.
Selected NCERT Exemplar Problems
Exercise 9.1
• Choose the correct answer from the given four options.
$$\textbf{1. A pole 6m high casts a shadow}\space2\sqrt{3}\\\textbf{m long on the ground, then the Sun’s elevation is}$$
(a) 60°
(b) 45°
(c) 30°
(d) 90°
Sol. (a) 60°
Explanation:
Let BC = 6 m be the height of the pole and AB $$=2\sqrt{3}$$be the length of the shadow on the ground. Let the Sun’s makes an angle θ on the ground.
In right angled ΔABC,
$$\text{tan}\space\theta=\frac{\text{BC}}{\text{AB}}=\frac{\text{6}}{2\sqrt{3}}\\\Rarr\space\text{tan}\theta=\sqrt{3}$$
⇒ θ = 60°.
Exercise 9.2
• State whether the following statements are true or false. Justify your answer.
1. If the length of the shadow of a tower is increasing, then the angle of elevation of a sun is also increasing.
Sol. False. We know, if the elevation moves towards the tower, it increases and if its elevation moves away the tower, if decreases. Hence, if the shadow of a tower is increasing, then the angle of elevation of a Sun is not increasing.
2. If a man standing on a platform, 3 m above the surface of a lake observes a cloud and its relfection in the lake, then the angle of elevation of the cloud is equal to the angle of depression of its reflection.
Sol. False. We know, if P is a point above the lake at a distance, d, then the reflection of the point in the lake would be at the same distance d. Also, the angle of elevation and depression from the surface of the lake is same.
Here, the man is standing on a platform 3 m above the surface, so its angle of elevation tothe cloud and angle of depression to the reflection of the cloud is not same.
3. The angle of elevation of the top of a tower is 30°. If the height of the tower is doubled, then the angle of elevation of its top will also be doubled.
Sol. False. Let height of the tower be h
then BC = h and let AB = x
In right asngled ΔABC,
$$\text{tan 30\degree}=\frac{h}{x}\\\Rarr\space\frac{1}{\sqrt{3}}=\frac{h}{x}\\\Rarr\space x=h\sqrt{3}$$
Suppose, if ∠BAC = θ and BC = 2h
$$\text{Then,}\space\text{tan}\space\theta=\frac{2h}{x}\\=\frac{2h}{h\sqrt{3}}=\frac{2}{\sqrt{3}}\\\Rarr\space\theta=\text{tan}^{\normalsize-1}\bigg(\frac{2}{\sqrt{3}}\bigg)\neq60\degree$$
4. 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.
Sol. True. Let BC = h be the height of the tower and A be the position of the observer makes an angle of elevation is θ.
Also, let AB = x
In right anglerd ΔABC,
$$\text{tan}\space\theta=\frac{h}{x}$$
Suppose, if height increase by 10%.
$$\text{i.e.,\space BC = h+}\frac{10}{100}h\\=\frac{11h}{10}$$
And if distance between observer and foot of the tower increase by10%.
$$\text{i.e., AB = x + }\frac{10}{100}x\\=\frac{11x}{10}\\\text{tan}\phi=\frac{11h/10}{11x/10}=\frac{h}{x}$$
= tan θ
⇒ φ = θ
Hence, again of elevation remains unchanged.
Exercise 9.3
1. Find the angle of elevation of the Sun, $$\textbf{when the shadow of a pole h m}\\\textbf{high is}\sqrt{3}\textbf{h m long.}$$
Sol. Let BC = h be the height of the tower and
$$\text{AB}=\sqrt{3}\space\text{h}\text{be the length of the shadow.}$$
Let the Sun’s elevation be ∠BAC = θ
In right angled ΔABC,
$$\text{tan}\theta=\frac{\text{BC}}{\text{AB}}\\=\frac{\text{h}}{\sqrt{3}\text{h}}=\frac{1}{\sqrt{3}}$$
⇒ θ = 30°
2. A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, find the height of the wall.
Sol. Let BC = h metre be the height of the wall and AC = 15 m be the length of the ladder. The ladder AC makes an angle of elevation ∠BAC = 60°.
In right angled ΔABC,
$$\text{sin 60\degree}=\frac{\text{AB}}{\text{AC}}\\\Rarr\space\frac{\sqrt{3}}{2}=\frac{h}{15}\\\Rarr\space h=\frac{15\sqrt{3}}{2}\text{m}$$
3. An observer 1.5 m tall is 20.5 m awayfrom a tower 22 m high. Determine the angle of elevation of the top of the tower from the eye of the observer.
Sol. Let BE = 22 m be the height of the tower and AD = 1.5 m be the length of the observer.
Also, AB = 20.5 m = DC
Let θ be the angle make by observer to the top of the tower.
In right angled ΔDCE,
$$\text{tan}\space\theta=\frac{\text{CE}}{\text{DC}}=\frac{\text{20.5}}{\text{\text{20.5}}}=1$$
⇒ θ = 45°
Exercise 9.4
1. The angle of elevation of the top of a tower from two points distant s and t from its foot are complementary.
$$\textbf{Prove that the height of the tower is}\sqrt{\textbf{st}}$$
Sol. Let CD = h metre be the height of the tower.
Let A and B be any two points on the ground such that they makes a distant s and t from the foot of the tower.
At points A and B, they makes an angle of elevation to the top of the lower are.
∠CAD = θ and ∠CBD = 90° – θ
(in complementary the sum of two angles are 90°).
In right angled ΔCAD.
$$\text{tan}\space\theta=\frac{\text{CD}}{\text{AC}}\\\Rarr\space\text{tan}\space\theta=\frac{h}{s}\\\space\text{...(i)}\\\text{And in right angled ΔCBD,}\\\text{tan(90\degree-}\theta)=\frac{\text{CD}}{\text{BC}}\\\Rarr\space\text{cot}\theta=\frac{h}{t}\space\text{...(ii)}$$
On multiplying eqs. (i) and (ii), we get
$$\text{tan}\theta×\text{cot}\theta=\frac{h}{s}×\frac{h}{t}\\\Rarr\space 1=\frac{h^{2}}{\text{st}}\Rarr h^{2}=st$$
On multiplying eqs. (i) and (ii), we get
$$\text{tan}\theta×\text{cot}\theta=\frac{h}{s}×\frac{h}{t}\\\Rarr\space 1=\frac{h^{2}}{st}\Rarr h^{2}=st$$
$$\Rarr\space h=\sqrt{\text{st}}\space\text{(Taking square root)}$$
Hence proved.
2. The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation is 30°, then when it is 60°. Find the height of the tower.
Sol. Let CD = h metre be the height of the tower. When the Sun makes an elevation 30° to 60°, the shadow of tower is AB = 50 m. And angles are ∠CAD = 30° = ∠CBD = 60°.
Also, let BC = x metre
In right angled ΔCAD,
$$\text{tan 30\degree}=\frac{\text{CD}}{\text{AC}}\\\Rarr\space\frac{1}{\sqrt{3}}=\frac{h}{\text{AB+BC}}\\\Rarr\space\frac{1}{\sqrt{3}}=\frac{h}{50+x}\\\Rarr\space 50+x=h\sqrt{3}\space\text{...(i)}$$
And in right angled ΔCBD,
$$\text{tan 60}\degree=\frac{\text{CD}}{\text{BC}}\\\Rarr\space\sqrt{3}=\frac{h}{x}\\\Rarr\space x=\frac{h}{\sqrt{3}}\\\text{Putting x}=\frac{h}{\sqrt{3}}\\\text{in eq. (i), we get}\\50+\frac{h}{\sqrt{3}}=h\sqrt{3}\\\Rarr\space h\bigg(\sqrt{3}-\frac{1}{\sqrt{3}}\bigg)=50$$
$$\Rarr\space h=\frac{\frac{50}{3-1}}{\sqrt{3}}\\\Rarr\space h=\frac{50\sqrt{3}}{2}=25\sqrt{3}\\\text{Hence, the height of the tower is}\\25\sqrt{3}\text{m.}$$
3. A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h. At a point on the plane, the angles of elevation of the bottom and the top of the flagstaff are α and b, respectively. Prove that the height of the tower is
$$\bigg(\frac{h\text{tan}\alpha}{\text{tan}\beta-\text{tan}\alpha}\bigg).$$
Sol. Let CD = H metre be the height of the tower and DE = h metre be the height of the flagstaff, which is standing on the tower.
Let A be a point on the plane, which makes an angle of β and α to the top and bottom of the flag.
Also, let AC = x m
In right angled ΔCAE,
$$\text{tan}\space\beta=\frac{\text{CD}}{\text{AC}}=\frac{\text{CD+DE}}{x}\\\Rarr\space\text{tan}\beta=\frac{\text{H+h}}{x}\space\text{...(i)}$$
In right angled ΔCAD,
$$\text{tan}\alpha=\frac{\text{CD}}{\text{AC}}\\\Rarr\space\text{tan}\alpha=\frac{\text{H}}{x}$$
⇒ x = H cot α
Putting x = H cot α in eq. (i), we get
$$\text{tan}\beta=\frac{\text{H+h}}{\text{H cot}\space\alpha}\\=\frac{(\text{H+h})}{\text{H}}\text{tan}\alpha$$
⇒ H(tan β – tan β) = h tan β
$$\Rarr\space\text{H}=\frac{\text{h tan}\space\alpha}{\text{tan}\beta-\text{tan}\alpha}$$
Hence proved.
4. The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 60° and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the distance between the two towers and also the height of the other tower.
Sol. Let BC = 30 m and AD = h metre be the height of the first and second tower.
From the foot of the towers B and A the angles of elevation are
∠DBA = 30°
and ∠BAC = 60°
Let distance between two towers are AB = x m
In right angled ΔABD,
$$\text{tan 30\degree}=\frac{\text{AD}}{\text{AB}}\\\Rarr\space\frac{1}{\sqrt{3}}=\frac{h}{x}\space\text{...(i)}$$
And in right angled ΔABC,
$$\text{tan 60}\degree=\frac{\text{BC}}{\text{AB}}\\\Rarr\space\sqrt{3}=\frac{30}{x}\\\Rarr\space x=\frac{30}{\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}}\\=\frac{30\sqrt{3}}{3}$$
$$\Rarr\space\text{x = 10}\sqrt{3}\text{m}\\\text{Putting x=10}\sqrt{3}\text{in eq. (i), we get}\\10\sqrt{3}=h\sqrt{3}\\\Rarr h=10\space m$$
Hence, the distance two towers are$$10\sqrt{3}\text{m}$$ and height of the other tower is 10 m.
5. From the top of a tower h metre high, the angle of depression of two objects which are in line with the foot of the tower are α and β (β > α). Find the distance between the two objects.
Sol. Let CD = h metre be the height of the tower. Let A and B be any two objects on the line. From top of the tower point D, the angle of depression of objects are ∠ODA = β and ∠ODB = β.
⇒ ∠DAC = α and ∠DBC = β (alternate angle)
Let distance between two objects AB = x m and BC = y m.
In right angled ΔCAD,
$$\text{tan}\space\alpha=\frac{\text{CD}}{\text{AC}}\\\Rarr\space\text{tan}\alpha=\frac{h}{\text{AB+BC}}\\\Rarr\text{tan}\alpha=\frac{h}{x+y}$$
⇒ x + y = h cot α ...(i)
In ΔCBD,
$$\text{tan}\space\beta=\frac{h}{y}$$
⇒ y = h cot β
Putting y = cot b in eq. (i), we get
x + h cot β = h cot α
⇒ x = h (cot α – cot β)
Hence, the distance between two objects is h(cot α – cot β).
6. A ladder rests against a vertical wall at a indination α to the horizontal. Its foot is pulled away from the wall through a distance p, so that its upper end slides a distance q down the wall and then the ladder makes an angle β to the horizontal.
$$\textbf{Show that}\space\frac{\textbf{p}}{\textbf{q}}=\frac{\textbf{cos}\beta-\textbf{cos}\alpha}{\textbf{sin}\alpha-\textbf{sin}\beta}.$$
Sol. Let BC = H be the height of the wall and AC = l metre be the length of the ladder which rests against a vertical wall at an angle ∠BAC = α. When the ladder pulled away from wall, its new position will be DE = l metre and AD = p metre.
Let AB = x and EC = q
In right angled ΔABC,
$$\text{sin}\alpha=\frac{\text{H}}{\text{l}}\\\text{and}\space\text{cos}\alpha=\frac{x}{l}$$
and in right angled ΔBDE,
$$\text{sin}\beta=\frac{\text{H-q}}{\text{I}}\\\text{and}\space\text{cos}\beta=\frac{p+x}{l}\\\text{RHS}=\frac{\text{cos}\beta-\text{cos}\alpha}{\text{cos}\alpha-\text{sin}\beta}\\=\frac{\frac{p+x}{l}-\frac{x}{l}}{\frac{H}{l}-\frac{\text{H-q}}{l}}\\=\frac{\text{p+x-x}}{\text{H-(H-q)}}=\frac{p}{q}\\=\text{LHS}$$
LHS = RHS Hence proved.
7. The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 45°. Finid the height of the tower.
Sol. Let BD = h metre be the height of the tower and A be a point on a ground.
At point A, the angle of elevation of the top of the tower is ∠BAD = 60° and E be a point on the tower which is 10 m above the point A.
Let AB = x m,
In right angled ΔBAD,
$$\text{tan 60}\degree=\frac{\text{BD}}{\text{AB}}\\\Rarr\space\sqrt{3}=\frac{h}{x}\\\Rarr\space x=\frac{h}{\sqrt{3}}\space\text{...(i)}$$
And in right angled ΔCED,
$$\text{tan 45}\degree=\frac{\text{CD}}{\text{CE}}\\\Rarr\space 1-\frac{h-10}{x}$$
⇒ x = h – 10
$$\Rarr\space\frac{h}{\sqrt{3}}=h-10\\\text{[From eq. (i)]}\\\Rarr\space 10=h\bigg(1-\frac{1}{\sqrt{3}}\bigg)\\\Rarr\space h=\frac{10\sqrt{3}}{(\sqrt{3}-1)}×\frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}\\=\frac{10\sqrt{3}}{2}(\sqrt{3}+1)\\=5(3+\sqrt{3})\text{m}\\\text{Hence, the height of the tower is}\\5(3+\sqrt{3})\text{m}.$$
8. A window of a house is h metre above the ground. From the window, the angles of elevation and depression of the top and the bottom of another house situated on the opposite side of the lane are found to be α and β, respectively. Prove that the height of the other house is h(1 + tan α cot β) m.
Sol. Let B be the position of the window.
∴ AB = h metre
Let CD = H metre be the height of another house which is situated on the opposite side of the lane. From the position of window B, the elevation and depression are
∠EBD = α
and ∠EBC = β = ∠BCA
(Alternate angle)
Draw a perpendicular line BE to the CD.
Let AC = BE = x
In right angled ΔEBD,
$$\text{tan}\space\alpha=\frac{\text{ED}}{\text{BE}}\\\Rarr\space\text{tan}\space\alpha=\frac{\text{H-h}}{x}$$
x = (H – h) cot α ...(i)
In right angled ΔABC,
$$\text{tan}\beta=\frac{\text{AB}}{\text{CA}}\\\Rarr\space\text{tan}\beta=\frac{h}{x}\\\Rarr\space\text{tan}\beta=\frac{h}{(\text{H-h})\text{cot}\alpha}$$
[From eq. (i)]
⇒ H – h = h tan β cot β
⇒ H = h(1 + tan α cot β)
Hence proved.
9. The tower window of a house is at a height of 2 m above the ground and its upper window is 4 m vertically above the lower window. At certain instant the angles of elevation of a balloon from these windows are observed to be 60° and 30°, respectively. Find the height of the balloon above the ground.
Sol. Let AG be the height of the balloon.
C and D be the position of the window.
Here, BC = 2 m, C = 4 m. At points C and D it makes an elevations are
∠ECD = 60° and ∠FDF = 30°
Draw a perpendicular line EC and FD onAG. Also, let CE = DF = x metre and GF = h metre.
In right angled ΔFDG,
$$\text{tan 30}\degree=\frac{\text{GF}}{\text{DF}}=\frac{h}{x}\\\Rarr\frac{1}{\sqrt{3}}=\frac{h}{x}\\\Rarr\space x=\sqrt{3}\space h\space\text{...(i)}$$
In right angled ΔECG,
$$\text{tan 60}\degree=\frac{\text{EG}}{\text{CE}}\\\Rarr\space\sqrt{3}=\frac{h+4}{x}\\\Rarr\space\sqrt{3}=\frac{h+4}{\sqrt{3h}}$$
[From eq.(i)]
3h = h + 4
⇒ 2h = 4
⇒ h = 2m
∴ Required height of the balloon
AG = AE + EF + FG = 2 + 4 + 2 = 8m.
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