# NCERT Solutions for Class 10 Maths Chapter 9 Exercise 9.1

Mathematics is an important subject for students in all facets of their lives, not just in school. Therefore, having a solid understanding of Mathematics will help them not just in school, but also in their daily lives. Class 10 Mathematics is extremely important since students study various important topics with many applications. By properly learning Mathematics and gaining high grades on board exams, students can lay a solid foundation for further studies. One of the most critical and essential subjects for students in Class 10 is Mathematics. It is sometimes regarded as the most difficult subject among students in Class 10. It calls for extensive practise with the problems as well as a deep understanding of the principles. Typically, students learn and retain every concept in the curriculum, but they do not practise answering questions enough. As a result, they are put at a disadvantage when sitting in the board exams. Therefore, it is crucial that they practise the questions in the NCERT textbook exercises. As these books are suggested by the CBSE, NCERT books serve as the primary source of instruction in the majority of schools. This increases the need for NCERT solutions. For senior classes, such as Class 10, NCERT solutions are crucial for board exam preparation. To give students plenty of practise before the board exam, Extramarks offers NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1 and other study modules.

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**NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry (Ex 9.1) H2 – Exercise 9.1**

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**Access NCERT Solutions for Class 10 Maths Chapter 9 – Trigonometry**

Prior to the board exams, students must have access to a large number of questions as well as well-explained solutions. The vast majority of question banks mainly depend on the NCERT textbook questions. The NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1 make answering questions easier. When students understand and learn quickly, they are much more confident in any examination.Students can also devote more time to all other elements of their studies whilst also managing to keep their minds cool. The NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1 completely answer all of the questions in Ex 9.1 Class 10. As a result, repeatedly practising the NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1 will enable students to answer such questions quickly and accurately in exams.

**NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Exercise 9.1**

The NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1 are an important resource for Class 10 students who want to get a head start on their preparation for the Class 10 board examinations in Mathematics. The CBSE recommends the NCERT-published Mathematics textbook for Class 10. Several other boards include the NCERT textbook in their curricula as well. Chapter 9 titled – Some Applications of Trigonometry is introduced to Class 10 students of the NCERT Mathematics book. Trigonometry is applied to find the heights and distances of various objects without actually measuring them. Trigonometry is used by astronomers to find the distances of planets and stars from the Earth. The NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1, along with the study material provided by Extramarks, cover all of the important concepts covered by the CBSE syllabus for the board examinations. The NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1 are developed by expert teachers to prepare students to effectively solve questions that appear in the CBSE board examinations. Practising the NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1 also facilitates students’ preparation for various competitive examinations like the Olympiad and others. To be thorough with the problems, students should practise the NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1 at least twice.

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**NCERT Solutions for Class 10 Maths PDFs **

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**Q.1 ** A circus artist is climbing 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 the following figure).

**Ans.**

\begin{array}{l}\text{In the given figure above, we have}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{sin}\text{\hspace{0.17em}}\text{30\xb0}=\frac{\text{AB}}{\text{AC}}=\frac{\text{AB}}{20}\\ \text{or}\frac{1}{2}=\frac{\text{AB}}{20}\\ \text{or AB}=\frac{20}{2}=10\\ \text{Therefore, height of the pole is 10 m}\text{.}\end{array}

**Q.2 ** A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground 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 the height of the tree.

**Ans.**

Let the tree AB bends at point C. Foot of the tree is at A and CB’ is the broken part of the tree i.e, BC = CB’.

$\begin{array}{l}\text{Now, in the above figure, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}30\xb0}=\frac{\text{AC}}{\text{AB\u2019}}=\frac{\text{AC}}{8}\\ \text{or}\frac{1}{\sqrt{3}}=\frac{\text{AC}}{8}\\ \text{or AC}=\frac{8}{\sqrt{3}}\\ \text{Also,}\\ \text{cos\hspace{0.17em}30\xb0}=\frac{\text{AB\u2019}}{\text{CB\u2019}}=\frac{\text{8}}{\text{CB\u2019}}\\ \text{or}\frac{\sqrt{3}}{2}=\frac{\text{8}}{\text{CB\u2019}}\\ \text{or CB\u2019}=\frac{16}{\sqrt{3}}\\ \text{Therefore, height of the tree}=\text{AC}+\text{CB\u2019}=\frac{8}{\sqrt{3}}+\frac{16}{\sqrt{3}}=\frac{24}{\sqrt{3}}=8\sqrt{3}\text{\hspace{0.17em}\hspace{0.17em}m}\end{array}$

**Q.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 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

**Ans.**

As per given information, we have the following figure.

$\begin{array}{l}\text{In}\mathrm{\Delta}\text{ABC, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}sin\hspace{0.17em}30\xb0}=\frac{\text{AB}}{\text{AC}}=\frac{1.5}{\text{AC}}\\ \text{or}\frac{1}{2}=\frac{1.5}{\text{AC}}\\ \text{or AC}=1.5\times 2=3\text{m}\\ \text{Therefore, length of the slide for the children below}\\ \text{the age of 5 years is}3\text{m.}\end{array}$

Also, for elder children, we have the following slide.

$\begin{array}{l}\text{In}\mathrm{\Delta}\text{PQR, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}sin\hspace{0.17em}}6\text{0\xb0}=\frac{\text{PQ}}{\text{PR}}=\frac{3}{\text{PR}}\\ \text{or}\frac{\sqrt{3}}{2}=\frac{3}{\text{PR}}\\ \text{or PR}=\frac{3\times 2}{\sqrt{3}}=2\sqrt{3}\text{m}\\ \text{Therefore, length of the slide for the elder children is}2\sqrt{3}\text{m.}\end{array}$

**Q.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.

**Ans.**

Let PQ is a tower with foot at Q and R is a point on the ground such that distance between Q and R is 30 m.

$\begin{array}{l}\text{In}\mathrm{\Delta}\text{PQR, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}30\xb0}=\frac{\text{PQ}}{\text{QR}}=\frac{\text{PQ}}{\text{30}}\\ \text{or}\frac{1}{\sqrt{3}}=\frac{\text{PQ}}{\text{30}}\\ \text{or PQ}=\frac{30}{\sqrt{3}}=10\sqrt{3}\text{m}\\ \text{Therefore, height of the tower is\hspace{0.33em}}10\sqrt{3}\text{m}.\end{array}$

**Q.5 ** A kite is 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.

**Ans.**

Let the kite is flying at point A and AC is the string of the kite tied at point C on the ground.

$\begin{array}{l}\text{In}\mathrm{\Delta}\text{ABC, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}sin\hspace{0.17em}}6\text{0\xb0}=\frac{\text{AB}}{\text{AC}}=\frac{60}{\text{AC}}\\ \text{or}\frac{\sqrt{3}}{2}=\frac{60}{\text{AC}}\\ \text{or AC}=\frac{60\times 2}{\sqrt{3}}=40\sqrt{3}\text{m}\\ \text{Therefore, length of the string is}40\sqrt{3}\text{m.}\end{array}$

**Q.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 he walks towards the building. Find the distance he walked towards the building.

**Ans.**

Let AB be a building of height 30 m. Let the boy is initially at D and then from D moves to point E. We have to find the distance DE

$\begin{array}{l}\text{In}\mathrm{\Delta}\text{ACE, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}6\text{0\xb0}=\frac{\text{AC}}{\text{EC}}=\frac{28.5}{\text{EC}}\\ \text{or}\sqrt{3}=\frac{28.5}{\text{EC}}\\ \text{or EC}=\frac{28.5}{\sqrt{3}}\text{m}\\ \text{Now,}\\ \text{In}\mathrm{\Delta}\text{ACD, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}3\text{0\xb0}=\frac{\text{AC}}{\text{CD}}=\frac{28.5}{\text{EC}+\text{DE}}\\ \text{or}\frac{1}{\sqrt{3}}=\frac{28.5}{\text{EC}+\text{DE}}\\ \text{or EC}+\text{DE}=28.5\sqrt{3}\text{}\\ \text{or DE}=28.5\sqrt{3}\text{}-\text{EC}=28.5\sqrt{3}-\frac{28.5}{\sqrt{3}}\\ \text{or DE}=\frac{28.5\times 3-28.5}{\sqrt{3}}=\frac{57}{\sqrt{3}}=\frac{19\times 3}{\sqrt{3}}=19\sqrt{3}\text{m}\\ \text{Therefore, the required distance is}19\sqrt{3}\text{m.}\end{array}$

**Q.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.

**Ans.**

Let BC be a building on which a transmission tower AB is fixed. Let D be a point on the ground.

$\begin{array}{l}\text{In}\mathrm{\Delta}\text{BCD, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}}45\text{\xb0}=\frac{\text{BC}}{\text{DC}}=\frac{20}{\text{DC}}\\ \text{or}1=\frac{20}{\text{DC}}\\ \text{or DC}=20\text{m}\\ \text{Now, in}\mathrm{\Delta}\text{\hspace{0.17em}ACD, we have}\\ \text{tan\hspace{0.17em}}60\text{\xb0}=\frac{\text{AC}}{\text{DC}}=\frac{\text{AC}}{20}\\ \text{or}\sqrt{3}=\frac{\text{AC}}{20}\\ \text{or AC}=20\sqrt{3}\text{}\\ \text{or AC}=\text{AB}+\text{BC}=20\sqrt{3}\text{}\\ \text{or AB}=20\sqrt{3}-\text{BC}=20\sqrt{3}-\text{20}=20(\sqrt{3}-1)\\ \text{Therefore, height of the tower is}20(\sqrt{3}-1)\text{m.}\end{array}$

**Q.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.

**Ans.**

Let AB be the statue on the pedestal BC. Let D be a point on the ground.

$\begin{array}{l}\text{In}\mathrm{\Delta}\text{BCD, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}}45\text{\xb0}=\frac{\text{BC}}{\text{DC}}\\ \text{or}1=\frac{\text{BC}}{\text{DC}}\\ \text{or DC}=\text{BC}\\ \text{Now, in}\mathrm{\Delta}\text{\hspace{0.17em}ACD, we have}\\ \text{tan\hspace{0.17em}}60\text{\xb0}=\frac{\text{AC}}{\text{DC}}=\frac{\text{AB}+\text{BC}}{\text{BC}}\\ \text{or}\sqrt{3}=\frac{\text{1.6}+\text{BC}}{\text{BC}}\\ \text{or}\sqrt{3}\text{\hspace{0.17em}BC}=\text{1.6}+\text{BC}\\ \text{or BC}(\sqrt{3}-1)\text{\hspace{0.17em}}=\text{1.6}\\ \text{or BC}=\frac{\text{1.6}}{(\sqrt{3}-1)}=\frac{\text{1.6}}{(\sqrt{3}-1)}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}=0.8(\sqrt{3}+1)\text{m}\\ \text{Therefore, height of the pedestal is}0.8(\sqrt{3}+1)\text{m.}\end{array}$

**Q.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.

**Ans.**

Let AB be the height of the building and CD be the height of the tower.

$\begin{array}{l}\text{In}\mathrm{\Delta}\text{BCD, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}}60\text{\xb0}=\frac{\text{CD}}{\text{BD}}\\ \text{or}\sqrt{3}=\frac{\text{50}}{\text{BD}}\\ \text{or BD}=\frac{\text{50}}{\sqrt{3}}\\ \text{Now, in}\mathrm{\Delta}\text{\hspace{0.17em}ABC, we have}\\ \text{tan\hspace{0.17em}}30\text{\xb0}=\frac{\text{AB}}{\text{BD}}\\ \text{or}\frac{1}{\sqrt{3}}=\frac{\text{AB}}{\frac{\text{50}}{\sqrt{3}}}\\ \text{or AB}=\frac{1}{\sqrt{3}}\times \frac{\text{50}}{\sqrt{3}}=\frac{\text{50}}{3}=16\frac{2}{3}\\ \text{Therefore, height of the building is}16\frac{2}{3}\text{m.}\end{array}$

**Q.10 ** Two poles of equal heights are standing opposite each other on either side of the road, which is 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.

**Ans.**

Let AB and CD be the two poles of equal height. Let O be a point on the road from where elevation angles are measured.

$\begin{array}{l}\text{In}\mathrm{\Delta}\text{\hspace{0.17em}ABO, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}}60\text{\xb0}=\frac{\text{AB}}{\text{BO}}\\ \text{or}\sqrt{3}=\frac{\text{AB}}{\text{BO}}\\ \text{or BO}=\frac{\text{AB}}{\sqrt{3}}\\ \text{Now, in}\mathrm{\Delta}\text{\hspace{0.17em}CDO, we have}\\ \text{tan\hspace{0.17em}}30\text{\xb0}=\frac{\text{CD}}{\text{OD}}\\ \text{or}\frac{1}{\sqrt{3}}=\frac{\text{CD}}{\text{OD}}=\frac{\text{CD}}{\text{BD}-\text{BO}}=\frac{\text{CD}}{\text{80}-\text{BO}}\\ \text{or CD}\sqrt{3}=\text{80}-\text{BO}=\text{80}-\frac{\text{AB}}{\sqrt{3}}\\ \text{or AB}\sqrt{3}=\text{80}-\frac{\text{AB}}{\sqrt{3}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}[\text{AB}=\text{CD as heights of the poles are equal]}\\ \text{or AB}\sqrt{3}+\frac{\text{AB}}{\sqrt{3}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{80}\\ \text{or}\frac{4\text{AB}}{\sqrt{3}}=80\\ \text{or AB}=20\sqrt{3}\\ \text{Therefore, height of each pole is}20\sqrt{3}\text{m.}\\ \text{BO}=\frac{\text{AB}}{\sqrt{3}}=\frac{20\sqrt{3}}{\sqrt{3}}=20\text{m}\\ \text{OD}=\text{BD}-\text{BO}=80-20=60\text{m}\\ \text{Therefore, the required distances are 20 m and 60 m.}\end{array}$

**Q.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 the following figure). Find the height of the tower and the width of the canal.

**Ans.**

We have the following figure.

$\begin{array}{l}\text{In \Delta ABD, we have}\\ \text{tan\hspace{0.17em}30\xb0 =}\frac{\text{AB}}{\text{DB}}\text{=}\frac{\text{AB}}{\text{BC+CD}}\\ \text{or}\frac{\text{1}}{\sqrt{\text{3}}}\text{=}\frac{\text{AB}}{\text{BC+20}}\\ \text{or BC+20 = AB}\sqrt{\text{3}}\\ \text{or BC = AB}\sqrt{\text{3}}-\text{20\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{(1)}\\ \text{Now,}\\ \text{In \Delta \hspace{0.17em}ABC, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}60\xb0 =}\frac{\text{AB}}{\text{BC}}\\ \text{or}\sqrt{\text{3}}\text{=}\frac{\text{AB}}{\text{BC}}\\ \text{or AB = BC}\sqrt{\text{3}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{(2)}\\ \text{From equations (1) and (2), we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}BC = AB}\sqrt{\text{3}}-\text{20}\\ \text{or BC = BC}\sqrt{\text{3}}\text{\hspace{0.17em}\xd7}\sqrt{\text{3}}-\text{20=3BC-20}\\ \text{or 3BC}-\text{BC = 20}\\ \text{or 2BC = 20}\\ \text{or BC = 10 m}\\ \text{On putting this value of BC in equation (2), we get}\\ \text{AB = 10}\sqrt{\text{3}}\text{m}\\ \text{Therefore, height of the tower is 10}\sqrt{\text{3}}\text{m and width of}\\ \text{the canal is 10 m.}\end{array}$

**Q.12 ** From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

**Ans.**

Let AB is a cable tower and CD is a building.

$\begin{array}{l}\text{We have, CD\hspace{0.33em}=\hspace{0.33em}BE\hspace{0.33em}=\hspace{0.33em}7 m, DB\hspace{0.33em}=\hspace{0.33em}CE}\\ \text{In \Delta \hspace{0.17em}CDB, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}45\xb0\hspace{0.33em}=\hspace{0.33em}}\frac{\text{CD}}{\text{DB}}\\ \text{or 1\hspace{0.33em}=\hspace{0.33em}}\frac{\text{7}}{\text{DB}}\\ \text{or DB =\hspace{0.33em}7 m}\\ \text{Now,}\\ \text{In \Delta \hspace{0.17em}ACE, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}60\xb0\hspace{0.33em}=\hspace{0.33em}}\frac{\text{AE}}{\text{CE}}\\ \text{or}\sqrt{\text{3}}\text{\hspace{0.33em}=\hspace{0.33em}}\frac{\text{AE}}{\text{CE}}\\ \text{or AE\hspace{0.33em}= CE}\sqrt{\text{3}}\text{\hspace{0.33em}= DB}\sqrt{\text{3}}\text{\hspace{0.33em}=\hspace{0.33em}7}\sqrt{\text{3}}\text{m\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}[CE\hspace{0.17em}= DB]}\\ \text{Therefore,}\\ \text{height of the tower\hspace{0.33em}=\hspace{0.33em}AB =\hspace{0.33em}AE+BE =\hspace{0.33em}7}\sqrt{\text{3}}\text{+7 =\hspace{0.33em}7(1+}\sqrt{\text{3}}\text{) m}\end{array}$

**Q.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.

**Ans.**

$\begin{array}{l}\text{Let AB be a lighthouse and ships be at points C and D.}\\ \text{It is given that AB = 75 m. We have to find the distance CD.}\\ \text{In \Delta \hspace{0.17em}ABC, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}45\xb0 =}\frac{\text{AB}}{\text{BC}}\\ \text{or 1 =}\frac{\text{AB}}{\text{BC}}\\ \text{or BC = AB = 75 m}\\ \text{Now,}\\ \text{In \Delta \hspace{0.17em}ABD, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}30\xb0 =}\frac{\text{AB}}{\text{DB}}\\ \text{or}\frac{\text{1}}{\sqrt{\text{3}}}\text{=}\frac{\text{75}}{\text{DB}}\\ \text{or DB = 75}\sqrt{\text{3}}\\ \text{or DB = BC+CD = 75}\sqrt{\text{3}}\\ \text{or CD = 75}\sqrt{\text{3}}-\text{BC = 75}\sqrt{\text{3}}-\text{75 = 75\hspace{0.17em}(}\sqrt{\text{3}}-\text{1)}\\ \text{Therefore,}\\ \text{distance between the two ships = 75\hspace{0.17em}(}\sqrt{\text{3}}-\text{1) m}\end{array}$

**Q.14 ** A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m 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 the following figure). Find the distance travelled by the balloon during the interval.

**Ans.**

$\begin{array}{l}\text{Let the balloon be initially at A and then it moves to B.}\\ \text{Let CD be the girl.}\\ \text{In \Delta \hspace{0.17em}ACE, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}60\xb0 =}\frac{\text{AE}}{\text{CE}}\\ \text{or}\sqrt{\text{3}}\text{=}\frac{\text{AE}}{\text{CE}}\\ \text{or CE =}\frac{\text{AE}}{\sqrt{\text{3}}}\text{=}\frac{\text{AF}-\text{EF}}{\sqrt{\text{3}}}\text{=}\frac{\text{88.2}-\text{1.2}}{\sqrt{\text{3}}}\text{=}\frac{\text{87}}{\sqrt{\text{3}}}\text{= 29}\sqrt{\text{3}}\\ \text{In \Delta \hspace{0.17em}BCG, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}30\xb0 =}\frac{\text{BG}}{\text{CG}}\\ \text{or}\frac{\text{1}}{\sqrt{\text{3}}}\text{=}\frac{\text{88.2}-\text{1.2}}{\text{CG}}\\ \text{or CG = 87}\sqrt{\text{3}}\text{m}\\ \text{Therefore,}\\ \text{distance travelled by the balloon = EG}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em} \hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}= CG-CE}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em} \hspace{0.33em}= 87}\sqrt{\text{3}}-\text{29}\sqrt{\text{3}}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em} = 58}\sqrt{\text{3}}\text{m}\end{array}$

**Q.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.

**Ans.**

$\begin{array}{l}\text{Let AB be a man and BC be a tower. Let the car is initially}\\ \text{at D and after 6 seconds it moves to point E.}\\ \text{In \Delta \hspace{0.17em}ACE, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}60\xb0 =}\frac{\text{AC}}{\text{CE}}\\ \text{or}\sqrt{\text{3}}\text{=}\frac{\text{AC}}{\text{CE}}\\ \text{or AC =}\sqrt{\text{3}}\text{\hspace{0.17em}CE}...\text{(1)}\\ \text{Now,}\\ \text{In \Delta \hspace{0.17em}ACD, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}30\xb0 =}\frac{\text{AC}}{\text{CD}}\\ \text{or}\frac{\text{1}}{\sqrt{\text{3}}}\text{=}\frac{\text{AC}}{\text{CD}}\\ \text{or CD = AC}\sqrt{\text{3}}\\ \text{or CD =}\sqrt{\text{3}}\text{\xd7}\sqrt{\text{3}}\text{\hspace{0.17em}CE = 3CE}\left[\text{from (1), AC =}\sqrt{\text{3}}\text{\hspace{0.17em}CE}\right]\\ \text{or DE+CE = 3CE}\\ \text{or DE = 2CE}\\ \text{or CE =}\frac{\text{DE}}{\text{2}}\\ \text{The car takes 6 seconds to cover the distance DE.}\\ \text{Therefore, it will take 3 seconds to cover the half of the}\\ \text{distance DE i.e., the car will take 3 seconds to reach the}\\ \text{foot of the tower from the point E.}\end{array}$

**Q.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.

**Ans.**

Let AB is a tower. Let C and D respectively are two points at a distance of 4 m and 9 m from the base of the tower.

$\begin{array}{l}\text{Let angles of elevation of the top of the tower from D be \theta}\\ \text{and from C be 90\xb0}-\text{\theta .}\\ \text{In \Delta \hspace{0.17em}ABC, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}(90\xb0}-\text{\theta )\hspace{0.33em}=\hspace{0.33em}}\frac{\text{AB}}{\text{BC}}\\ \text{or cot\theta \hspace{0.33em}=\hspace{0.33em}}\frac{\text{AB}}{\text{4}}\\ \text{or AB\hspace{0.33em}=\hspace{0.33em}4cot\theta}...\text{(1)}\\ \text{Now,}\\ \text{In \Delta \hspace{0.17em}ABD, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}\theta \hspace{0.33em}=\hspace{0.33em}}\frac{\text{AB}}{\text{BD}}\\ \text{or \hspace{0.17em}tan\hspace{0.17em}\theta \hspace{0.33em}=\hspace{0.33em}}\frac{\text{AB}}{\text{9}}\\ \text{or \hspace{0.17em}\hspace{0.17em}AB\hspace{0.33em}=\hspace{0.33em}9 tan\theta}...\text{(2)}\\ \text{From (1) and (2), we have}\\ \text{9 tan\theta \hspace{0.33em}=\hspace{0.33em}4cot\theta}\\ \text{or \hspace{0.17em}\hspace{0.17em}}\frac{\text{tan\theta}}{\text{cot\theta}}\text{\hspace{0.33em}=\hspace{0.33em}}\frac{\text{4}}{\text{9}}\\ {\text{or \hspace{0.17em}\hspace{0.17em}tan}}^{\text{2}}\text{\theta \hspace{0.33em}=\hspace{0.33em}}\frac{\text{4}}{\text{9}}\\ \text{or \hspace{0.17em}\hspace{0.17em}tan\theta \hspace{0.33em}=\hspace{0.33em}}\frac{\text{2}}{\text{3}}\\ \text{On putting this value of tan\theta in (2), we get}\\ \text{AB\hspace{0.33em}=\hspace{0.33em}9 tan\theta \hspace{0.33em}=\hspace{0.33em}9\xd7}\frac{\text{2}}{\text{3}}\text{\hspace{0.33em}=\hspace{0.33em}6 m}\\ \text{Therefore, the height of the tower is 6 m.}\end{array}$

## FAQs (Frequently Asked Questions)

### 1. What is the number of questions in the NCERT Solutions For Class 10 Maths Chapter 9 Exercise 9.1?

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