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NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

by anand
September 22, 2019
in 10th Class
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NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry – Here are all the NCERT solutions for Class 10 Maths Chapter 8. This solution contains questions, answers, images, explanations of the complete Chapter 8 titled Introduction to Trigonometry of Maths taught in Class 10. If you are a student of Class 10 who is using NCERT Textbook to study Maths, then you must come across Chapter 8 Introduction to Trigonometry. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry in one place.

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NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry

Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Maths for Class 10 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

For a better understanding of this chapter, you should also see summary of Chapter 8 Introduction to Trigonometry , Maths, Class 10.

Class 10
Subject Maths
Book Mathematics
Chapter Number 8
Chapter Name

Introduction to Trigonometry

NCERT Solutions Class 10 Maths chapter 8 Introduction to Trigonometry

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Class 10, Maths chapter 8, Introduction to Trigonometry solutions are given below in PDF format. You can view them online or download PDF file for future use.

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NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

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Question & Answer

Q.1: In \(\triangle\) ABC right angled at B, AB=24 cm, BC=7 m. Determine
(i) \(\sin A, \cos A\)
(ii) \(\sin C, \cos C\)

Ans : \(\begin{array}{l}{\text { Applying Pythagoras theorem for } \Delta A B C, \text { we obtain }} \\ {A C^{2}=A B^{2}+B C^{2}} \\ {=(24 \mathrm{cm})^{2}+(7 \mathrm{cm})^{2}} \\ {=(576+49) \mathrm{cm}^{2}} \\ {=625 \mathrm{cm}^{2}} \\ {\therefore A C=\sqrt{625} \mathrm{cm}=25 \mathrm{cm}}\end{array}\) \(\sin A=\frac{\text { Side opposite to } \angle A}{\text { Hypotenuse }}=\frac{B C}{A C}\) \(=\frac{7}{25}\) \(\cos A=\frac{\text { Side adjacent to } \angle A}{\text { Hypotenuse }}=\frac{A B}{A C}=\frac{24}{25}\) (ii) \(\begin{array}{l}{ \sin \mathrm{C}=\frac{\text { Side opposite to } \angle \mathrm{C}}{\text { Hypotenuse }}=\frac{\mathrm{AB}}{\mathrm{AC}}} \\ {=\frac{24}{25}}\end{array}\) \(\begin{array}{l}{\cos \mathrm{C}=\frac{\text { Side adjacent to } \angle \mathrm{C}}{\mathrm{Hypotenuse}}=\frac{\mathrm{BC}}{\mathrm{AC}}} \\ {=\frac{7}{25}}\end{array}\)

Q.2: In the given figure find tan P - cot R.
 

Ans : \(\begin{array}{l}{\text { Applying Pythagoras theorem for } \Delta P Q R, \text { we obtain }} \\ {P R^{2}=P Q^{2}+Q R^{2}} \\ {(13 \mathrm{cm})^{2}=(12 \mathrm{cm})^{2}+Q R^{2}} \\ {169 \mathrm{cm}^{2}=144 \mathrm{cm}^{2}+Q R^{2}} \\ {25 \mathrm{cm}^{2}=Q R^{2}} \\ {Q R=5 \mathrm{cm}}\end{array}\) \(\begin{aligned} \tan \mathbf{P} &=\frac{\text { Side opposite to } \angle P}{\text { Side adjacent to } \angle P}=\frac{Q R}{P Q} \\ &=\frac{5}{12} \\ \cot R &=\frac{\text { Side adjacent to } \angle R}{\text { Side opposite to } \angle R}=\frac{Q R}{P Q} \\ &=\frac{5}{12} \end{aligned}\) \(\tan \mathrm{P}-\cot \mathrm{R}=\frac{5}{12}-\frac{5}{12}=0\)

Q.3: If \(\sin A=\frac{3}{4}, \text { calculate cos } A \text { and } \tan A.\)

Ans : Let\(\Delta A B C \text { be a right-angled triangle, right-angled at point } B\) Given that, \(\begin{array}{l}{\sin A=\frac{3}{4}} \\ {\frac{B C}{A C}=\frac{3}{4}}\end{array}\) Let BC be 3k . Therefore, AC will be 4k , where k is a positive integer. Applying Pythagoras theorem in \(\triangle ABC \), we obtain \(\begin{array}{l}{A C^{2}=A B^{2}+B C^{2}} \\ {(4 k)^{2}=A B^{2}+(3 k)^{2}} \\ {16 k^{2}-9 k^{2}=A B^{2}} \\ {7 k^{2}=A B^{2}} \\ {A B=\sqrt{7} k}\end{array}\) \(\begin{aligned} \cos A &=\frac{\text { Side adjacent to } \angle A}{\text { Hypotent } t 0 \angle A} \\ &=\frac{A B}{A C}=\frac{\sqrt{7 k}}{4 k}=\frac{\sqrt{7}}{4} \\ \tan A &=\frac{\text { Side opposite to } \angle A}{\text { Side adjacent to } \angle A} \\ &=\frac{B C}{A B}=\frac{3 k}{\sqrt{7} k}=\frac{3}{\sqrt{7}} \end{aligned}\)

Q.4: Given \(15 \cot A=8 . \text { Find } \sin A \text { and } \sec A.\)

Ans : Consider a right-angled triangle, right-angled at B. Given that, \(\begin{array}{l}{\sin A=\frac{3}{4}} \\ {\frac{B C}{A C}=\frac{3}{4}}\end{array}\) Let BC be 3k. Therefore, AC will be 4k, where k is a positive integer. Applying Pythagoras theorem in \(\triangle \mathrm{ABC}\) , we obtain \(\begin{array}{l}{\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}} \\ {(4 k)^{2}=\mathrm{AB}^{2}+(3 \mathrm{k})^{2}} \\ {16 \mathrm{k}^{2}-\mathrm{gk}^{2}=\mathrm{AB}^{2}} \\ {7 k^{2}=\mathrm{AB}^{2}} \\ {\mathrm{AB}=\sqrt{7} k}\end{array}\) \(\begin{aligned} \cos A &=\frac{\text { Side adjacent to } \angle A}{\text { Hypotenuse }} \\ &=\frac{A B}{A C}=\frac{\sqrt{7 k}}{4 k}=\frac{\sqrt{7}}{4} \end{aligned}\) \(\begin{aligned} \tan A &=\frac{\text { Side opposite to } \angle A}{\text { Side adjacent to } \angle A} \\ &=\frac{B C}{A B}=\frac{3 k}{\sqrt{7} k}=\frac{3}{\sqrt{7}} \end{aligned}\) \(\begin{aligned} \cot A &=\frac{\text { Side adjacent to } \angle A}{\text { Side opposite to } \angle A} \\ &=\frac{A B}{B C} \end{aligned}\) \(\begin{array}{c}{\text { It is given that, }} \\ {\text { cot } A=\frac{8}{15}} \\ {\frac{A B}{B C}=\frac{8}{15}}\end{array}\) Let AB be 8k.Therefore, 3C will be 15k, where k is a positive integer. Applying Pythagoras theorem in \(\triangle\)ABC, we obtain \(\begin{array}{l}{\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}} \\ {=(8 k)^{2}+(15 k)^{2}} \\ {=64 k^{2}+225 \mathrm{k}^{2}} \\ {=289 \mathrm{k}^{2}} \\ {\mathrm{AC}=17 k}\end{array}\) \(\begin{aligned} \sin A &=\frac{\text { Side opposite to } \angle A}{\text { Hypotenuse }}=\frac{B C}{A C} \\ &=\frac{15 k}{17 k}=\frac{15}{17} \\ \sec A &=\frac{\text { Hypotenuse }}{\text { Side adjacent to } \angle A} \\ &=\frac{A C}{A B}=\frac{17}{8} \end{aligned}\)

Q.5: Given \(\sec \theta=\frac{13}{12},\) calculate all other trigonometric ratios.

Ans : Consider a right-angle \(\triangle \mathrm{ABC}, \text { right-angled at point } \mathrm{B}\) \(\begin{array}{l}{\sec \theta=\frac{\text { Hypotenuse }}{\text { Side adjacent to } \angle \theta}} \\ {\frac{13}{12}=\frac{\mathrm{AC}}{\mathrm{AB}}}\end{array}\) If AC is 13k, AB will be 12k, where k is a positive integer. Applying Pythagoras theorem in AABC, we obtain \(\begin{array}{l}{(A C)^{2}=(A B)^{2}+(B C)^{2}} \\ {(13 k)^{2}=(12 k)^{2}+(B C)^{2}} \\ {169 k^{2}=144 k^{2}+B C^{2}} \\ {25 k^{2}=B C^{2}}\end{array}\) BC = 5k \(\sin \theta=\frac{\text { Side opposite to } \angle \theta}{\text { Hypotenuse }}=\frac{B C}{A C}=\frac{5 k}{13 k}=\frac{5}{13}\) \(\cos \theta=\frac{\text { Side adjacent to } \angle \theta}{\text { Hypotenuse }}=\frac{A B}{A C}=\frac{12 k}{13 k}=\frac{12}{13}\) \(\tan \theta=\frac{\text { Side opposite to } \angle \theta}{\text { Side adjacent to } \angle \theta}=\frac{B C}{A B}=\frac{5 k}{12 k}=\frac{5}{12}\) \(\cot \theta=\frac{\text { Side adjacent to } \angle \theta}{\text { Side opposite to } \angle \theta}=\frac{A B}{B C}=\frac{12 k}{5 k}=\frac{12}{5}\) \(cosec \ \theta=\frac{\text { Hypotenuse }}{\text { Side opposite to } \angle \theta}=\frac{A C}{B C}=\frac{13 k}{5 k}=\frac{13}{5}\)

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