NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

Ans : \(\begin{array}{l}{\text { (i) } 25^{\circ}} \\ {\text { We know that } 180^{\circ}=\pi \text { radian }} \\ {\therefore 25^{\circ}=\frac{\pi}{180} \times 25 \text { radian }=\frac{5 \pi}{36} \text { radian }}\end{array}\)
\(\begin{array}{l}{\text { (ii) }-47^{\circ} 30^{\prime}} \\ {-47^{\circ} 30^{\prime}=-47 \frac{1}{2}} \\ {=\frac{-95}{2}}\end{array}\)
Since \(180^{\circ}=\pi\) radian
\(\begin{array}{l}{\text { (iii) } 240^{\circ}} \\ {\text { We know that } 180^{\circ}=\pi \text { radian }} \\ {\therefore 240^{\circ}=\frac{\pi}{180} \times 240 \text { radian }=\frac{4}{3} \pi \text { radian }}\end{array}\)
\(\begin{array}{l}{\text { (iv) } 520^{\circ}} \\ {\text { We know that } 180^{\circ}=\pi \text { radian }} \\ {\therefore 520^{\circ}=\frac{\pi}{180} \times 520 \text { radian }=\frac{26 \pi}{9} \text { radian }}\end{array}\) 

Ans : \(\begin{array}{l}{\text { (i) } \frac{11}{16}} \\ {\text { We know that } \pi \text { radian }=180^{\circ}}\end{array}\)
\(\begin{aligned} \therefore \frac{11}{16} \operatorname{radain} &=\frac{180}{\pi} \times \frac{11}{16} \text { degree }=\frac{45 \times 11}{\pi \times 4} \text { deg ree } \\ &=\frac{45 \times 11 \times 7}{22 \times 4} \text { deg ree }=\frac{315}{8} \text { degree } \end{aligned}\)
\(\begin{array}{l}{=39 \frac{3}{8} \text { deg ree }} \\ {=39^{\circ}+\frac{3 \times 60}{8} \text { min utes } \quad\left[1^{\circ}=60^{\circ}\right]} \\ {=39^{\circ}+22^{\prime}+\frac{1}{2} \text { min utes }} \\ {=39^{\circ} 22^{\prime} 30^{\prime \prime}} & {\left[1^{\prime}=60^{\prime \prime}\right]}\end{array}\)
\(\begin{array}{l}{(ii)-4} \\ {\text { We know that } \pi \text { radian }=180^{\circ}} \\ {-4 \text { radian }=\frac{180}{\pi} \times(-4) \text { deg ree }=\frac{180 \times 7(-4)}{22} \text { degree }} \\ {=\frac{-2520}{11} \text { deg ree }=-229 \frac{1}{11} \text { degree }}\end{array}\)
\(=-229^{\circ}+\frac{1 \times 60}{11} \text { min utes } \quad\left[1^{\circ}=60^{\circ}\right]\)
\(=-229^{\circ} 5^{12} 27^{n} \quad \quad\left[1^{\prime}=60^{\prime \prime}\right]\)
\(\begin{array}{l}{\text { (iii) } \frac{5 \pi}{3}} \\ {\text { We know that } \pi \text { radlan }=180^{\circ}} \\ {\therefore \frac{5 \pi}{3} \text { radian }=\frac{180}{\pi} \times \frac{5 \pi}{3} \text { degree }=300^{\circ}}\end{array}\)
\(\begin{array}{l}{\text { (iv) } \frac{7 \pi}{6}} \\ {\text { We know that } \pi \text { radian }=180^{\circ}} \\ {\therefore \frac{7 \pi}{6} \text { radian }=\frac{180}{\pi} \times \frac{7 \pi}{6}=210^{\circ}}\end{array}\) 

Ans : Number of revolutions made by the wheel in 1 minute = 360
\(\therefore \) Number of revolutions made by the wheel in 1 second = \(\frac{360}{60}=6\) 
Hence, In 6 complete revolutions, it will turn an angle  of  \(6 \times 2 \pi\) radian, i.e., 12 \(\pi \) radian  
Thus, In one second, the wheel turns and angle of 12 \(\pi \) radian. 

Ans : We know that in a circle of radius r unit, if an arc of length l unit subtends and angle \(\theta \) radian at the centre, then
\(\theta=\frac{1}{r}\)
Therefore, for r = 100 cm, l = 22 cm, we have 
\(\begin{aligned} \theta &=\frac{22}{100} \text { radian }=\frac{180}{\pi} \times \frac{22}{100} \text { deg ree }=\frac{180 \times 7 \times 22}{22 \times 100} \text { deg ree } \\ &=\frac{126}{10} \text { deg ree }=12 \frac{3}{5} \text { deg ree }=12^{\circ} 36^{\circ} \quad\left[1^{\circ}=60^{\prime}\right] \end{aligned}\)
Thus, the required angle is \(12^{\circ} 36^{\prime}\). 

Ans : Diameter of the circle = 40 cm
Radius of  the circle = \(\frac{40}{2} \mathrm{cm}=20 \mathrm{cm}\)
Let AB be a chord (length = 20 cm) of the circle.

\(\begin{array}{l}{\text { In } \triangle O A B, O A=O B=\text { Radius of circle }=20 \mathrm{cm}} \\ {\text { Also, } A B=20 \mathrm{cm}} \\ {\text { Thus, } \triangle O A B \text { is an equilateral triangle. }}\end{array}\)
\(\therefore \theta=60^{\circ}=\frac{\pi}{3}\)
We know that in a circle of radius r unit, if an arc of length l unit subtends an angle \(\theta\) radian at the centre, then \(\theta=\frac{l}{r}\)
\(\frac{\pi}{3}=\frac{\widetilde{A B}}{20} \Rightarrow \widetilde{A B}=\frac{20 \pi}{3} \mathrm{cm}\)
Thus, the length of the minor arc of the chord is \(\frac{20 \pi}{3} \mathrm{cm}\).