NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

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) 
\(\begin{array}{l}{ \sin \mathrm{C}=\frac{\text { Side opposite to } \angle \mathrm{C}}{\text { Hypotenuse }}=\frac{\mathrm{AB}}{\mathrm{AC}}} \\ {=\frac{24}{25}}\end{array}\)
\(\begin{array}{l}{\cos \mathrm{C}=\frac{\text { Side adjacent to } \angle \mathrm{C}}{\mathrm{Hypotenuse}}=\frac{\mathrm{BC}}{\mathrm{AC}}} \\ {=\frac{7}{25}}\end{array}\) 

Ans : \(\begin{array}{l}{\text { Applying Pythagoras theorem for } \Delta P Q R, \text { we obtain }} \\ {P R^{2}=P Q^{2}+Q R^{2}} \\ {(13 \mathrm{cm})^{2}=(12 \mathrm{cm})^{2}+Q R^{2}} \\ {169 \mathrm{cm}^{2}=144 \mathrm{cm}^{2}+Q R^{2}} \\ {25 \mathrm{cm}^{2}=Q R^{2}} \\ {Q R=5 \mathrm{cm}}\end{array}\)

\(\begin{aligned} \tan \mathbf{P} &=\frac{\text { Side opposite to } \angle P}{\text { Side adjacent to } \angle P}=\frac{Q R}{P Q} \\ &=\frac{5}{12} \\ \cot R &=\frac{\text { Side adjacent to } \angle R}{\text { Side opposite to } \angle R}=\frac{Q R}{P Q} \\ &=\frac{5}{12} \end{aligned}\)
\(\tan \mathrm{P}-\cot \mathrm{R}=\frac{5}{12}-\frac{5}{12}=0\) 

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
\(\begin{array}{l}{A C^{2}=A B^{2}+B C^{2}} \\ {(4 k)^{2}=A B^{2}+(3 k)^{2}} \\ {16 k^{2}-9 k^{2}=A B^{2}} \\ {7 k^{2}=A B^{2}} \\ {A B=\sqrt{7} k}\end{array}\)
\(\begin{aligned} \cos A &=\frac{\text { Side adjacent to } \angle A}{\text { Hypotent } t 0 \angle A} \\ &=\frac{A B}{A C}=\frac{\sqrt{7 k}}{4 k}=\frac{\sqrt{7}}{4} \\ \tan A &=\frac{\text { Side opposite to } \angle A}{\text { Side adjacent to } \angle A} \\ &=\frac{B C}{A B}=\frac{3 k}{\sqrt{7} k}=\frac{3}{\sqrt{7}} \end{aligned}\) 

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
\(\begin{array}{l}{\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}} \\ {=(8 k)^{2}+(15 k)^{2}} \\ {=64 k^{2}+225 \mathrm{k}^{2}} \\ {=289 \mathrm{k}^{2}} \\ {\mathrm{AC}=17 k}\end{array}\)
\(\begin{aligned} \sin A &=\frac{\text { Side opposite to } \angle A}{\text { Hypotenuse }}=\frac{B C}{A C} \\ &=\frac{15 k}{17 k}=\frac{15}{17} \\ \sec A &=\frac{\text { Hypotenuse }}{\text { Side adjacent to } \angle A} \\ &=\frac{A C}{A B}=\frac{17}{8} \end{aligned}\) 

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
\(\sin \theta=\frac{\text { Side opposite to } \angle \theta}{\text { Hypotenuse }}=\frac{B C}{A C}=\frac{5 k}{13 k}=\frac{5}{13}\)
\(\cos \theta=\frac{\text { Side adjacent to } \angle \theta}{\text { Hypotenuse }}=\frac{A B}{A C}=\frac{12 k}{13 k}=\frac{12}{13}\)
\(\tan \theta=\frac{\text { Side opposite to } \angle \theta}{\text { Side adjacent to } \angle \theta}=\frac{B C}{A B}=\frac{5 k}{12 k}=\frac{5}{12}\)
\(\cot \theta=\frac{\text { Side adjacent to } \angle \theta}{\text { Side opposite to } \angle \theta}=\frac{A B}{B C}=\frac{12 k}{5 k}=\frac{12}{5}\)
\(cosec \ \theta=\frac{\text { Hypotenuse }}{\text { Side opposite to } \angle \theta}=\frac{A C}{B C}=\frac{13 k}{5 k}=\frac{13}{5}\)