Ch. - 3 Trigonometric Functions

Ch. - 3 Trigonometric Functions

Chapter 3 Ex.3.1 Question 1

Find the radian measures corresponding to the following degree measures:

(i) \(25^\circ\)

(ii) \(-47^{\circ}\) \(30^\prime\)

(iii) \(240^{\circ}\)

(iv) \(520^{\circ}\)

Solution

We know that \(180^\circ = \pi \) radian

Therefore,

(i) \(25^\circ\)

\[\begin{align}25^\circ &= \frac{\pi }{180} \times 25 \quad \text{radian }\\&= \frac{5\pi }{36} \quad \text{radian }\end{align}\]

(ii) \(-47^{\circ}\) \(30^\prime\)

\[\begin{align} - 47^\circ 30\prime &= - \left( {47\frac{1}{2}} \right)^\circ = - \left( {\frac{{95}}{2}} \right)^\circ\\- \left( {\frac{{95}}{2}} \right)^\circ &= \frac{\pi }{{180}} \times - \left( {\frac{{95}}{2}} \right)\quad \text{radian}\\&= - \frac{19\pi }{72} \quad \text{ radian}\end{align}\]

(iii) \(240^{\circ}\)

\[\begin{align}240^\circ &= \frac{\pi }{{180}} \times 240\quad \text{radian}\\&= \frac{{3\pi }}{4} \quad \text{radian}\end{align}\]

(iv) \(520^{\circ}\)

\[\begin{align}520^\circ &= \frac{\pi }{180} \times 520 \quad \text{radian}\\ &= \frac{26\pi }{9} \quad \text{ radian} \end{align}\]

Chapter 3 Ex.3.1 Question 2

Find the degree measures corresponding to the following radian measures \(\begin{align}\left( {\rm{Use}\;\pi = \frac{22}{7}} \right).\end{align}\)

(i) \(\frac{11}{16}\)

(ii) \(-4 \)

(iii) \(\frac{5\pi }{3}\)

(iv) \(\frac{7\pi }{6}\)

Solution

We know that \(180^\circ = \pi \) radian

Therefore,

(i) \(\frac{11}{16}\)

\[\begin{align}\frac{11}{16} \; \text{radian} &= \frac{180^\circ }{\pi } \times \frac{11}{16}\\ &=180\times \frac{7}{22}\times \frac{11}{16}\deg \\ & =\frac{315}{8}\deg \\ & =39\frac{3}{8}\deg \\ & =39\deg +\frac{3}{8}\times 60\min \qquad \quad\left[ \because 1{}^\circ =6{0}' \right] \\ & =39\deg +22\frac{1}{2}\min \\ & =39\deg +22\min +\frac{60}{2}\sec \qquad \left[ \because {1}'=6{0}'' \right] \\ & =39{}^\circ 2{2}'3{0}'' \end{align}\]

(ii) \(-4 \)

\[\begin{align}-4\; \text{radian} &= \frac{180^\circ }{\pi } \times \left( { - 4} \right)\\ & =180\times \frac{7}{22}\times \left( -4 \right)\deg \\ & =-\frac{2520}{11}\deg \\ & =-229\frac{1}{11}\deg \\ & =-229\deg +\frac{60}{11}\min \qquad \qquad \left[ \because 1{}^\circ =6{0}' \right] \\ & =-22\deg +5\frac{5}{11}\min \\ & =-22\deg +5\min +\frac{5}{11}\times 60\sec \qquad \left[ \because {1}'=6{0}'' \right] \\ & =-22{}^\circ {5}'2{7}'' \\ \end{align}\]

(iii) \(\frac{5\pi }{3}\)

\[\begin{align}\frac{{5\pi }}{3}\;\text{radian} &= \frac{{180^\circ }}{\pi } \times \frac{{5\pi }}{3}\\& = 300^\circ \end{align}\]

(iv) \(\frac{7\pi }{6}\)

\[\begin{align}\frac{7\pi }{6}\text{ radian}&= \frac{180^\circ }{\pi } \times \frac{7\pi }{6}\\ &= 210^\circ \end{align}\]

Chapter 3 Ex.3.1 Question 3

A wheel makes \(360\) revolutions in one minute. Through how many radians does it turn in one second?

Solution

A wheel makes \(360\) revolutions in \(1\) minute (\(60\) seconds)

Therefore,

Number of revolutions made by the wheel in \(1\) second \( = \frac{{360}}{{60}} = 6\)

In one complete revolution, the wheel turns an angle of \(2\pi \) radians

Hence, in 6 complete revolutions, it will turn an angle of \(6 \times 2\pi = 12\pi \) radians

Thus, in one second, the wheel turns an angle of \(12\pi \) radians.

Chapter 3 Ex.3.1 Question 4

Find the degree measure of the angle subtended at the centre of a circle of radius \(100\, \rm{cm}\) by an arc of length \(22\,\rm{cm}\).\(\begin{align}\left( {\rm{Use}\;\pi = \frac{22}{7}} \right) \end{align}\)

Solution

As we know that if in a circle of radius \(r,\) an arc of length l subtends an angle of \(\rm{\theta}\) radians,

Then \(l = r\theta \)

Therefore,

\[\begin{align}\theta &= \frac{l}{r}\text { radian} \\\theta &= \frac{\text{22 cm}}{\text{100 cm}}\text { radian}\\ \theta &=\frac{11}{50}\times \frac{180}{\pi }\deg \\ & =\frac{11}{50}\times 180\times \frac{7}{22}\deg \\ & =\frac{63}{5}\deg \\ & =12\frac{3}{5}\deg \\ & =12\deg +\frac{3}{5}\times 60\min \qquad \left[ \because 1{}^\circ =6{0}' \right] \\ & =12{}^\circ 3{6}'\end{align}\]

Thus, the required angle is \(12^\circ 36'\)

Chapter 3 Ex.3.1 Question 5

In a circle of diameter \(40\,\rm{ cm},\) the length of a chord is \(20 \,\rm{cm}.\) Find the length of minor arc of the chord.

Solution

Diameter of the circle\( = 40\,\rm{ cm}\)

Therefore, Radius of the circle \(r = \frac{\rm{40\,cm}}{2} = 20\,\rm{cm}\)

Let \(AB\) be a chord of length \(20\,\rm{ cm}\) of the circle.

In \(\Delta AOB\)

\(AB = 20cm\)

\(OA = OB = r = 20\,\rm{cm}\)

Hence, \(\Delta AOB\) is an equilateral triangle

Thus, \(\theta = 60^\circ \) or \(\theta = \frac{\pi }{3}\) radian

As we know that if in a circle of radius \(r,\) an arc of length \(l\) subtends an angle of \(θ\) radians,

Then \(l = r\theta \)

Therefore,

\[\begin{align} l&=r\theta \\ {AB} &=20\,\rm{cm}\times \frac{\pi }{3} \\ & =\frac{20\pi }{3}\,\rm{cm}\end{align}\]

Hence, the length of the minor arc of the chord is \(\frac{20\pi }{3}\,\rm{cm}\).

Chapter 3 Ex.3.1 Question 6

If in two circles, arcs of the same length subtend angles \(60^\circ\) and \(75^\circ\) at the centre, find the ratio of their radii.

Solution

Let the radii of the two circles be \(r\) and \(R.\)

Let an arc of length l subtend an angle of \(60^\circ\) at the centre of the circle of radius \(r,\) and \(75^\circ\) at the centre of the circle of radius \(R.\)

Now,

\(60^\circ = \frac{\pi }{3}\) radian and \(75^\circ = \frac{{5\pi }}{{12}}\) radian

As we know that if in a circle of radius \(r,\) an arc of length \(l\) subtends an angle of \(θ\) radians,

Then \(l = r\theta \)

Therefore,

\(\begin{align}l &= r \times \frac{\pi }{3}\\ &= \frac{{\pi r}}{3}\end{align}\) and \(\begin{align}l &= R \times \frac{5\pi }{12}\\ &= \frac{5\pi R}{12}\end{align}\)

Thus,

\[\begin{align}\frac{{\pi r}}{3} &= \frac{{5\pi R}}{{12}}\\\frac{r}{R} &= \frac{5}{4}\\r:R &= 5:4\end{align}\]

Hence, the ratio of their radii is \(5:4\).

Chapter 3 Ex.3.1 Question 7

Find the angle in radian through which a pendulum swings if its length is \(75\,\rm{ cm}\) and the tip describes an arc of length

(i) \(10\,\rm{cm}\)

(ii) \(15\, \rm{cm}\)

(iii) \(21\,\rm{ cm}\)

Solution

As we know that if in a circle of radius \(r,\) an arc of length \(l\) subtends an angle of \(θ\) radians,

Then \(l = r\theta \)

(i) Radius, \(r = 75 \,\rm{cm}\) and length of the arc, \(l = 10 \rm{cm}\)

\[\begin{align}\theta &= \frac{l}{r}\\ &= \frac{\rm{10cm}}{\rm{75\,cm}}\\ &= \frac{2}{15}\end{align}\]

Thus, \(\theta = \frac{2}{{15}}\) radian

(ii) Radius, \(r = 75\,\rm{cm}\) and length of the arc, \(l = 15\,\rm{cm}\)

\[\begin{align}\theta &= \frac{l}{r}\\ &= \frac{\rm{15\,cm}}{\rm{75\,cm}}\\ &= \frac{1}{5}\end{align}\]

Thus, \(\theta = \frac{1}{5}\) radian

(iii) Radius, \(r = 75\,\rm{cm}\) and length of the arc, \(l = 21\,\rm{cm}\)

\[\begin{align}\theta &= \frac{l}{r}\\ &= \frac{\rm{21\,cm}}{\rm{75\,cm}}\\ &= \frac{7}{{25}}\end{align}\]

Thus, \(\theta = \frac{7}{25}\) radian

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