Want to know how sailors navigated the oceans or how astronomers measured stars? The secret lies in trigonometry. Applications of trigonometry prove that math is the ladder we climb to measure the surrounding heights. In the last chapter, students learned about trigonometric ratios and identities. This chapter primarily covers the concepts of heights and distances, angles of elevation, angles of depression, and the line of sight. The primary benefit of using NCERT Solutions for Class 10 is that they offer clear explanations and accurate answers, facilitating a deeper understanding. Many teachers recommend NCERT Solutions because they closely match the exam pattern.
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Heights and distances are an essential part of Trigonometry, as learning these concepts thoroughly will not only help in class 10 but also in future classes and competitive exams. Trigonometry makes heights and distances easy to measure, demonstrating how math can connect the earth with the sky. These NCERT Solutions for Class 10 Maths are trustworthy and reliable, as they are created by subject matter experts at Careers360, making them an essential resource for exam preparation. For details about the syllabus, revisions, and to download PDFs of NCERT solutions, refer to the NCERT article.
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The NCERT Solutions for Class 10 Maths Chapter 9 have been prepared by Careers360 experts to make learning simpler and to help you score better in exams. A downloadable PDF is available — click the link below to access it.
NCERT Class 10 Maths Chapter 9 Some Applications of Trigonometry question answers with detailed explanations are provided below.
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Some Applications of Trigonometry Class 10 Question Answers |
Answer:
Given that,
The length of the rope (AC) = 20 m. and $\angle ACB$ $=30°$
Let the height of the pole (AB) be $h$
So, in the right triangle $\Delta ABC$
By using the sine rule
$\sin \theta = \frac{P}{H} =\frac{AB}{AC}$
$\sin 30° =\frac{h}{20}$
$h =10$ m.
Hence, the height of the pole is 10 m.
Answer:
Suppose DB is a tree, and the AD is the broken height of the tree, which touches the ground at C.
Given that,
$\angle ACB = 30°$ , BC = 8 m
let AB = $x$ m and AD = $y$ m
So, AD+AB = DB = $x+y$
In right angle triangle $\Delta ABC$ ,
$\tan \theta = \frac{P}{B}=\frac{x}{8}$
$\tan 30° =\frac{x}{8}=\frac{1}{\sqrt{3}}$
So, the value of $x$ = $\frac{8}{\sqrt{3}}$
Similarily,
$\cos 30° = \frac{BC}{AC} = \frac{8}{y}$
the value of $y$ is $\frac{16}{\sqrt{3}}$
So, the total height of the tree is-
$x+y=\frac{24}{\sqrt{3}}=8\sqrt{3}= 8 (1.732) = 13.856$ m (approx)
Answer:
Suppose $x$ m is the length of slides for children below 5 years, and the length of slides for older children is $y$ m.
Given that,
AF = 1.5 m, BC = 3 m, $\angle AEF = 30°$ and $\angle BDC = 60°$
In triangle $\Delta$ EAF,
$\sin \theta = \frac{AF}{EF} = \frac{1.5}{x}$
$\sin 30° = \frac{1.5}{x}$
The value of $x$ is 3 m.
Similarily in $\Delta$ CDB,
$\sin 60° = \frac{3}{y}$
$\frac{\sqrt{3}}{2}= \frac{3}{y}$
the value of $y$ is $2\sqrt{3}$ = 2(1.732) = 3.468
Hence, the length of the slide is for children below 5 years is 3 m, and for the elder children is 3.468 m.
Answer:
Let the height of the tower AB be $h$, and the angle of elevation from the ground at point C be $\angle ACB = 30°$
According to the question,
In the right triangle $\Delta ABC$ ,
$\tan \theta = \frac{AB}{BC} = \frac{h}{30}$
$\tan 30° =\frac{1}{\sqrt{3}}=\frac{h}{30}$
the value of $h$ is $10\sqrt{3}$ = 10(1.732) = 17.32 m
Thus, the height of the tower is 17.32 m
Answer:
Given that,
The length of AB = 60 m and the inclination of the string with the ground at point C is $\angle ACB = 60°$.
Let the length of the string AC be $l$.
According to the question,
In right triangle $\Delta$ CBA,
$\sin 60° = \frac{AB}{AC} = \frac{60}{l}$
$\frac{\sqrt{3}}{2} = \frac{60}{l}$
The value of length of the string ( $l$ ) is $40\sqrt{3}$ = 40(1.732) = 69.28 m
Hence, the length of the string is 69.28 m.
Answer:
Given that,
The height of the tallboy (DC) is 1.5 m, and the height of the building (AB) is 30 m.
$\angle ADF = 30°$ and $\angle AEF = 60°$
According to the question,
In right triangle AFD,
$\\\Rightarrow \tan 30°=\frac{AF}{DF} = \frac{28.5}{DF}$
$\Rightarrow \frac{1}{\sqrt{3}} = \frac{28.5}{DF}$
So, DF = $28.5\sqrt{3}$
In right angle triangle $\Delta AFE$,
$\tan 60° =\frac{AF}{FE}=\frac{28.5}{EF}$
$⇒\sqrt{3}=\frac{28.5}{EF}$
⇒ EF = $9.5\sqrt{3}$
So, distance walked by the boy towards the building = DF - EF = $19\sqrt{3}$
Answer:
Suppose BC = $h$ is the height of the transmission tower, and AB is the height of the building, and AD is the distance between the building and the observer point (D).
We have,
AB = 20 m, BC = $h$ m and AD = $x$ m
$\angle CDA = 60°$ and $\angle BDA = 45°$
According to the question,
In triangle $\Delta$ BDA,
$\tan 45° = \frac{AB}{AD}=\frac{20}{x}$
So, $x$ = 20 m
Again,
In triangle $\Delta$ CAD,
$\\\Rightarrow \tan 60° = \frac{AB+BC}{AD}=\frac{20+h}{20}$
$\Rightarrow \sqrt{3}= 1+\frac{h}{20}$
$\Rightarrow h=20(\sqrt{3}-1)$
$\Rightarrow h=20(0.732) = 14.64 m$
Answer- The height of the tower is 14.64 m
Answer:
Let the height of the pedestal be $h$ m, and the height of the statue be 1.6 m.
The angle of elevation of the top of the statue and top of the pedestal is ($\angle DCB=$ $60°$) and ($\angle ACB=$ $45°$ ) respectively.
Now,
In triangle $\Delta ABC$ ,
$\tan 45° =1 =\frac{AB}{BC}=\frac{h}{BC}$
therefore, BC = $h$ m
In triangle $\Delta CBD$ ,
$\\\Rightarrow \tan 60° = \frac{BD}{BC}=\frac{h+1.6}{h}$
$\Rightarrow \sqrt{3}= 1+\frac{1.6}{h}$
the value of $h$ is $0.8(\sqrt{3}+1)$ m
Hence the height of the pedestal is $0.8(\sqrt{3}+1)$ m
Answer:
It is given that the height of the tower (AB) is 50 m. $\angle AQB = 30°$ and $\angle PBQ = 60°$
Let the height of the building be $h$ m
According to the question,
In triangle PBQ,
$\tan 60° = \frac{PQ}{BQ} = \frac{50}{BQ}$
$⇒\sqrt{3} = \frac{50}{BQ}$
$⇒ BQ = \frac{50}{\sqrt{3}}$ .......................(i)
In triangle ABQ,
$\tan 30° = \frac{h}{BQ}$
${BQ}=h\sqrt{3}$ .........................(ii)
On equating equations (i) and (ii), we get,
$\frac{50}{\sqrt{3}}=h\sqrt{3}$
therefore, $h$ = $\frac{50}3$ = 16.66 m = height of the building.
Answer:
Given that,
The height of both poles is equal, DC = AB. The angles of elevation of the top of the poles are $\angle DEC=30°$ and $\angle AEB=60°$ respectively.
Let the height of the poles be $h$ m and CE = $x$ and BE = 80 - $x$
According to the question,
In triangle DEC,
$\\\Rightarrow \tan 30° = \frac{DC}{CE} = \frac{h}{x}$
$\Rightarrow \frac{1}{\sqrt{3}}= \frac{h}{x}$
$\Rightarrow x=h\sqrt{3}$ ..............(i)
In triangle AEB,
$\\\Rightarrow \tan 60° = \frac{AB}{BE}=\frac{h}{80-x}$
$\Rightarrow \sqrt{3}=\frac{h}{80-x}$
$\Rightarrow x=80 - \frac{h}{\sqrt{3}}$ ..................(ii)
On equating eq (i) and eq (ii), we get
$\begin{aligned} & \sqrt{3} h=80-\frac{h}{\sqrt{3}} \\ & \frac{h}{\sqrt{3}}=20\end{aligned}$
So, $x$ = 60 m
Hence, the height of both poles is ( $h=20\sqrt{3}$ )m, and the position of the point is at 60 m from the pole CD and 20 m from the pole AB.
Answer:
Suppose the $h$ is the height of the tower AB and BC = $x$ m
It is given that the width of CD is 20 m,
According to the question,
In triangle $\Delta ADB$ ,
$
\begin{aligned}
& \Rightarrow \tan 30^{\circ}=\frac{A B}{20+x}=\frac{h}{20+x} \\
& \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{20+x} \\
& \Rightarrow 20+x=h \sqrt{3} \\
& \Rightarrow x=h \sqrt{3}-20----(i)
\end{aligned}
$
In triangle ACB,
$\\\Rightarrow \tan 60° = \frac{h}{x}=\sqrt{3}$
$\Rightarrow x= \frac{h}{\sqrt{3}}$ .............(ii)
On equating equations (i) and (ii), we get:
$h\sqrt{3}-20= \frac{h}{\sqrt{3}}$
from here we can calculate the value of $h=10\sqrt{3}= 10 (1.732) = 17.32\: m$ and the width of the canal is 10 m.
Answer:
Let the height of the cable tower be (AB = $h+7$ )m
Given,
The height of the building is 7 m, and the angle of elevation of the top of the tower $\angle ACE = 60°$, the angle of depression of its foot $\angle BCE = 45°$.
According to the question,
In triangle $\Delta DBC$ ,
$\\\tan 45° = \frac{CD}{BD} = \frac{7}{BD} = 1\\ BD =7\ m$
since DB = CE = 7 m
In triangle $\Delta ACE$ ,
$\\\tan 60° = \frac{h}{CE}=\frac{h}{7}=\sqrt{3}$
$ \therefore h = 7\sqrt{3}\ m$
Thus, the total height of the tower equal to $h+7$ $=7(1+\sqrt{3}) m$
Answer:
Given that,
The height of the lighthouse (AB) is 75 m above sea level. And the angle of depression of two different ships are $\angle ADB = 30°$ and $\angle ACB = 45°$ respectively
Let the distance between the two ships be $x$ m.
According to the question,
In triangle $\Delta ADB$ ,
$\tan 30° = \frac{AB}{BD}=\frac{75}{x+y} = \frac{1}{\sqrt{3}}$
$\therefore x+y = 75 \sqrt{3}$ .............(i)
In triangle $\Delta ACB$ ,
$\tan 45° = 1 =\frac{75}{BC}=\frac{75}{y}$
$\therefore y =75\ m$ .............(ii)
From equations (i) and (ii), we get;
$x = 75(\sqrt{3}-1)=75(0.732)$
$x = 54.9\simeq 55\ m$
Hence, the distance between the two ships is approximately 55 m.
Answer:
Given that,
The height of the girl is 1.2 m. The height of the balloon from the ground is 88.2 m, and the angle of elevation of the balloon from the eye of the girl at any instant is ( $\angle ACB =60°$ ), and after some time, $\angle DCE =30°$.
Let the $x$ distance travelled by the balloon from position A to position D during the interval.
AB = ED = 88.2 – 1.2 = 87 m
Now, In triangle $\Delta BCA$ ,
$\\\tan 60° = \sqrt{3}=\frac{AB}{BC}=\frac{87}{BC}\\ $
$\therefore BC = 29\sqrt{3}$
In triangle $\Delta DCE$ ,
$\\\tan 30° = \frac{1}{\sqrt{3}}=\frac{DE}{CE}=\frac{87}{CE}$
$ \therefore CE = 87\sqrt{3}$
Thus, the distance travelled by the balloon from position A to D
$= CE - BC =87\sqrt{3}-29\sqrt{3} = 58\sqrt{3}$ m
Answer:
Let $h$ be the height of the tower (DC), and the speed of the car be $x\ ms^{-1}$.
Therefore, the distance (AB) covered by the car in 6 seconds is 6 $x$ m. Let $t$ be the time required to reach the foot of the tower. So, BC = $x$ $t$
According to the question,
In triangle $\Delta DAC$ ,
$\\\tan 30° = \frac{1}{\sqrt{3}}=\frac{h}{6x+xt}$
$⇒ x(6+t) = h\sqrt{3}$ ..........................(i)
In triangle $\Delta BCD$ ,
$\\\tan 60° = \sqrt{3} = \frac{h}{xt}\\ $
$\therefore h = 3xt$ ...................(ii)
Putting the value of $h$ in equation (i), we get,
$\\x(6+t) = (\sqrt{3}.\sqrt{3})xt\\$
$⇒ 6x +xt = 3xt\\ $
$⇒6x = 2xt$
$⇒t = 3$
Hence, from the point, the B car takes 3 seconds to reach the foot of the tower.
Answer:
Let the height of the tower be $h$ m.
we have PB = 4m and QB = 9 m
Suppose $\angle BQA = \theta$ , so $\angle APB =90- \theta$
According to the question,
In triangle $\Delta ABQ$ ,
$\\\tan \theta = \frac{h}{9}\\ $
$\therefore h = 9 \tan \theta$ ..............(i)
In triangle $\Delta ABP$ ,
$\\\tan (90-\theta)=\cot \theta = \frac{h}{4}\\$
$ \therefore h = 4\cot \theta$ .....................(ii)
Multiplying equations (i) and (ii), we get
$\\h^2 = 36\\ $
$\Rightarrow h = 6 m$
Hence, the height of the tower is 6 m.
NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry (Exercise 9.1) are provided below.
Question 1:
From the top of an upright pole 17.75 m high, the angle of elevation of the top of an upright tower was 60°. If the tower was 57.75 m tall, how far away (in m) from the foot of the pole was the foot of the tower?
Answer:
We have to find the value of $x$.
In $\triangle ABC,$
$\tan60° = \frac{AC}{BC}$
We know, $CE=BD$
⇒ $AC=AE-CE$
⇒ $AC=57.75-17.75$
⇒ $AC=40$
In $\triangle ABC,$
$\tan60° = \frac{40}{x}$
⇒ $\sqrt3=\frac{40}{x}$
⇒ $x=\frac{40}{\sqrt3}$
⇒ $x=\frac{40\sqrt3}{3}$ m
Hence, the correct answer is $\frac{40\sqrt3}{3}$.
Question 2:
The tops of two poles of heights 18 m and 30.5 m are connected by a wire. If the wire makes an angle of $30^{\circ}$ with the horizontal, what is the length (in m) of the wire?
Answer:
Here, $\mathrm{CD}=\mathrm{EB}=18 \mathrm{~m}$ and $\mathrm{CE}=\mathrm{DB}$
So, $\mathrm{AE}=30.5-18=12.5 \mathrm{~m}$
$\triangle A E C$ is a right-angled triangle at $E$.
$ \sin 30^{\circ}=\frac{A E} {A C}$
$ \Rightarrow \frac {1 }{2}=\frac {12.5}{ A C}$
$ \Rightarrow A C=25$
Hence, the correct answer is 25.
Question 3:
A 20 m long ladder rests against a wall so that the angle between the ladder and the wall is $30^{\circ}$. How far (in m) is the base of the ladder from the wall?
Answer:
Let AC ladder and AB be the wall.
$ACB =90^\circ -30^\circ=60^\circ$
$\cos 60^\circ=\frac {BC}{AC}$
$⇒\frac {1}{2}=\frac {BC}{20}$
$⇒BC =10\;m$
Hence, the correct answer is 10 m.
Question 4:
Exactly midway between the foot of two towers P and Q , the angles of elevation of their tops are $45^{\circ}$ and $60^{\circ}$, respectively. The ratio of the heights of $P$ and $Q$ is:
Answer:
Let the height of tower $P = h_1$, and the height of tower $Q = h_2. $
Let the distance between the two towers be $2d. $
The observer is located exactly midway, i.e., at a distance $d$ from each tower.
Using the tangent function for tower $P: $
$\tan 45^\circ = \frac{h_1}{d} \Rightarrow 1 = \frac{h_1}{d} \Rightarrow h_1 = d. $
Using the tangent function for tower $Q: $
$\tan 60^\circ = \frac{h_2}{d} \Rightarrow \sqrt{3} = \frac{h_2}{d} \Rightarrow h_2 = \sqrt{3} d. $
The required ratio of heights is:$\frac{h_1}{h_2} = \frac{d}{\sqrt{3} d} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}= \sqrt{3} $
Hence, the correct answer is $\sqrt{3}$.
Question 5:
The length of the shadow on the ground of a tall tree of height $30 \;m$ is $10 \sqrt{3} \mathrm{~m}$. What is the angle (in degrees) of elevation of the sun?
Answer:
Length $=30 \mathrm{~m}$ (Tower)
Base $=10 \sqrt{3} \mathrm{~m}$ (Shadow)
Let the angle of elevation of the sun be $\theta$
$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$
$⇒\tan \theta=\frac{30}{10 \sqrt{3}}$
$⇒\tan \theta=\frac{3}{\sqrt{3}} $
$⇒ \tan \theta=\sqrt{3}$
So, $\tan \theta=\tan 60^{\circ}$
So, $\theta=60^{\circ}$
Hence, the correct answer is $60^{\circ}$.
Students will explore the following topics in NCERT Class 10 Maths Chapter 9 Some Applications of Trigonometry:
Line of Sight - The Line of Sight is the line formed by our vision as it passes through an item when we look at it.
Horizontal Line - The distance between the observer and the object is measured by a horizontal line.
The angle of Elevation:
The angle formed by the line of sight to the top of the item and the horizontal line is called an angle of elevation.
It is above the horizontal line, i.e., when we gaze up at the item, we make an angle of elevation.
The angle of Depression:
When the spectator must look down to perceive the item, an angle of depression is formed.
When the horizontal line is above the angle, the angle of depression is formed between it and the line of sight.
Case 1:
In this case, we can observe the following:
Height of a tower, hill, or building
Distance of an object from the foot of the tower, hill, or building and sometimes the shadow of it
The angle of elevation or the angle of depression
Any two of the above three parameters will be provided in the question. This type of problem can be solved using the formulas given below.
In the right triangle $ABC$,
$\begin{aligned} & \sin \theta=\frac{\text { Opposite } }{ \text { Hypotenuse }}=\frac{\mathrm{AB} }{ \mathrm{AC}} \\ & \cos \theta=\frac{\text { Adjacent } }{ \text { Hypotenuse }}=\frac{\mathrm{BC} }{ \mathrm{AC}} \\ & \tan \theta=\frac{\text { Opposite } }{ \text { Adjacent }}=\frac{\mathrm{AB} }{\mathrm{BC}}\end{aligned}$
Case 2:
In this case, we can deal with different illustrations. One of the commonly solved problems is about the movement of an observer. If the observer moves toward objects like a tower, building, hill, etc., then the angle of elevation increases. The angle of elevation decreases when the observer moves away from the object. Here, the distance moved by the observer can be found using the formula given below:
In the right triangle given below, $d$ is the distance between $C$ and $D$.
$d = h(\cot x - \cot y)$
Case 3:
There is another case where two different situations happen at the same time. In this case, we get similar triangles with the same angle of elevation or angle of depression. These types of problems can be solved with the help of formulas related to similar triangles.
In the right triangle $ABC, DE || AB,$
Here, triangles $ABC$ and $EDC$ are similar.
Using Thales' or BPT theorem, we can write the ratio of sides as:
$\frac{AB}{ED} = \frac{BC}{DC}$
Trigonometry is not just theory, as it helps us solve real-life problems too. This chapter shows us how to use trigonometric ratios to find the heights and distances of different objects. These Class 10 Maths Chapter 9 Some Applications of Trigonometry question answers help students understand how maths connects with the real world. Here are some more points on why these question answers are important:
For students' preparation, Careers360 has gathered all Class 10 Maths NCERT solutions here for quick and convenient access.
Students must check the NCERT Exemplar solutions for class 10 of the Mathematics and Science Subjects. The following links will help them in this process.
Students can check the latest syllabus and some reference books from the following links.
Frequently Asked Questions (FAQs)
Some Applications of Trigonometry Class 10 question answers are based on Heights and Distances, where students learn how to apply trigonometric formulas to practical situations.
Yes. When drawn correctly from the same horizontal line, the angle of elevation from one point is equal to the angle of depression from the other.
Usually, metres (m) and centimetres (cm) are used, but both quantities must be in the same unit before calculation.
The angle of elevation is the angle formed when we look up at an object from the ground.
The angle of depression is the angle formed when we look down at an object from a higher point.
Many educational platforms, such as Careers360, offer free downloadable PDFs of NCERT Some Applications of Trigonometry Class 10 Solutions. Students can find the free downloadable PDF in this article itself.
Class 10 Chapter 9 of the maths textbook contains problems on height and distance, solvable using these steps.
The easiest way to understand trigonometric ratios in the NCERT solutions class 10th maths chapter 9 is:
In the Some Applications of Trigonometry class 10 questions with solutions, the following important formulas should be remembered:
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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is
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Option 2)
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Option 4)
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An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range
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Option 2)
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| Option 3)
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Option 4)
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A particle is projected at 600 to the horizontal with a kinetic energy . The kinetic energy at the highest point
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Option 2)
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| Option 3)
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Option 4)
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In the reaction,
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Option 2)
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Option 4)
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How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?
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0.02 |
Option 2)
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