Q: Convert $\frac{11}{16}$ radians into degrees.

We use the relation Degree measure = Radian measure $\times \frac{180}{\pi}$. Using $\pi = \frac{22}{7}$, we get $\frac{11}{16} \times \frac{180}{\pi} = \frac{11}{16} \times \frac{180 \times 7}{22} = \frac{315}{8}^{\circ} = 39\frac{3}{8}^{\circ}$. Since $\frac{3}{8}^{\circ} = \frac{3}{8} \times 60'

Q: Find the degree measure of the angle subtended at the centre of a circle of radius $100$ cm by an arc of length $22$ cm (use $\pi = \frac{22}{7}$).

We know that $\theta = \frac{l}{r}$. Here, $l = 22$ cm and $r = 100$ cm. So, $\theta = \frac{22}{100}$ radians. Converting this to degrees: $\theta = \frac{22}{100} \times \frac{180}{\pi} = \frac{22}{100} \times \frac{180 \times 7}{22} = \frac{63}{5}^{\circ} = 12\frac{3}{5}^{\circ} = 12^{\circ} \lef

Q: What is the value of $\sin\left(-\frac{11\pi}{3}\right)$?

We know that $\sin(-\theta) = -\sin(\theta)$. So, $\sin\left(-\frac{11\pi}{3}\right) = -\sin\left(\frac{11\pi}{3}\right)$. We can write $\frac{11\pi}{3} = 4\pi - \frac{\pi}{3}$. Thus, $-\sin\left(4\pi - \frac{\pi}{3}\right) = -\left(-\sin\left(\frac{\pi}{3}\right)\right) = \sin\left(\frac{\pi}{3}\ri

Question 1

Convert $40^{\circ} 20'$ into radian measure.

Accepted Answer

We know that $1^{\circ} = 60'$. Therefore, $20' = \frac{20}{60}^{\circ} = \left(\frac{1}{3}ight)^{\circ}$. Thus, $40^{\circ} 20' = 40\frac{1}{3}^{\circ} = \frac{121}{3}^{\circ}$. To convert degrees to radians, we multiply by $\frac{\pi}{180}$. So, $\frac{121}{3} 	imes \frac{\pi}{180} = \frac{121\

Question 2

Find the radian measure corresponding to $300^{\circ}$.

Accepted Answer

To convert degrees to radians, multiply the degree measure by $\frac{\pi}{180}$. Therefore, $300^{\circ} = 300 	imes \frac{\pi}{180} = \frac{5\pi}{3}$ radians.

Question 3

Convert $\frac{11}{16}$ radians into degrees.

Accepted Answer

We use the relation Degree measure = Radian measure $	imes \frac{180}{\pi}$. Using $\pi = \frac{22}{7}$, we get $\frac{11}{16} 	imes \frac{180}{\pi} = \frac{11}{16} 	imes \frac{180 	imes 7}{22} = \frac{315}{8}^{\circ} = 39\frac{3}{8}^{\circ}$. Since $\frac{3}{8}^{\circ} = \frac{3}{8} 	imes 60'

Question 4

Find the degree measure of the angle subtended at the centre of a circle of radius $100$ cm by an arc of length $22$ cm (use $\pi = \frac{22}{7}$).

Accepted Answer

We know that $	heta = \frac{l}{r}$. Here, $l = 22$ cm and $r = 100$ cm. So, $	heta = \frac{22}{100}$ radians. Converting this to degrees: $	heta = \frac{22}{100} 	imes \frac{180}{\pi} = \frac{22}{100} 	imes \frac{180 	imes 7}{22} = \frac{63}{5}^{\circ} = 12\frac{3}{5}^{\circ} = 12^{\circ} \lef

Question 5

What is the value of $\sin\left(-\frac{11\pi}{3}\right)$?

Accepted Answer

We know that $\sin(-	heta) = -\sin(	heta)$. So, $\sin\left(-\frac{11\pi}{3}ight) = -\sin\left(\frac{11\pi}{3}ight)$. We can write $\frac{11\pi}{3} = 4\pi - \frac{\pi}{3}$. Thus, $-\sin\left(4\pi - \frac{\pi}{3}ight) = -\left(-\sin\left(\frac{\pi}{3}ight)ight) = \sin\left(\frac{\pi}{3}i

Trigonometric Functions — CBSE Class 11 Mathematics Important Questions & Revision Notes