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.
NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry
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| Class | 10 |
| Subject | Maths |
| Book | Mathematics |
| Chapter Number | 8 |
| Chapter Name |
Introduction to Trigonometry |
NCERT Solutions Class 10 Maths chapter 8 Introduction to Trigonometry
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.
Introduction to Trigonometry Download
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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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