TRIGONOMETRIC FUNCTIONS-Exercise 3.1
Maths - Exercise
Q1. Find the radian measures corresponding to the following degree measures:
(i) 25°
(ii) – 47°30′
(iii) 240°
(iv) 520°
Solution
To convert degree measures into radian measures, we use the standard relation \(180^\circ = \pi\) radians. Accordingly, each degree measure is multiplied by \(\dfrac{\pi}{180}\) to obtain the corresponding radian measure.
\((i)\quad 25^\circ\) \[ \begin{aligned} 25^\circ &= \dfrac{\pi}{180} \times 25^\circ \\ &= \dfrac{25\pi}{180} \\ &= \dfrac{5\pi}{36} \\[6pt] \end{aligned}\] \((ii)\quad -47^\circ 30'\) \[ \begin{aligned}&= -\dfrac{\pi}{180} \times 47.5^\circ \\ \quad -47^\circ 30' &= -\dfrac{47.5\pi}{180} \\ &= -\dfrac{95\pi}{360} \\ &= -\dfrac{19\pi}{72} \\[6pt] \end{aligned}\] \((iii)\quad 240^\circ\) \[\begin{aligned}\quad 240^\circ&= \dfrac{\pi}{180} \times 240^\circ \\ &= \dfrac{240\pi}{180} \\ &= \dfrac{4\pi}{3} \\[6pt] \end{aligned}\] \((iv)\quad 520^\circ\) \[\begin{aligned} \quad 520^\circ &= \dfrac{\pi}{180} \times 520^\circ \\ &= \dfrac{520\pi}{180} \\ &= \dfrac{26\pi}{9} \end{aligned} \]
Hence, the radian measures corresponding to the given degree measures are \(\dfrac{5\pi}{36}\), \(-\dfrac{19\pi}{72}\), \(\dfrac{4\pi}{3}\), and \(\dfrac{26\pi}{9}\) respectively.
Q2. Find the degree measures corresponding to the following radian measures. Use \(\pi=\dfrac{22}{7}\)
(i) \(\dfrac{11}{16}\)
(ii) -4
(iii) \(\dfrac{5\pi}{3}\)
(iv) \(\dfrac{7\pi}{6}\)
Solution
To convert radian measures into degree measures, we use the relation \(\pi \text{ radians} = 180^\circ\). Since \(\pi = \dfrac{22}{7}\), each given radian measure is multiplied by \(\dfrac{180}{\pi} = \dfrac{180 \times 7}{22}\) to obtain the corresponding degree measure.
\((i)\quad \dfrac{11}{16}\) \[ \begin{aligned} \quad \dfrac{11}{16} &= \dfrac{180}{\pi} \times \dfrac{11}{16} \\ &= \dfrac{180 \times 7}{22} \times \dfrac{11}{16} \\ &= \dfrac{90 \times 7}{16} \\ &= \dfrac{630}{16} \\ &= 39.375^\circ \\ &= 39^\circ + 0.375 \times 60' \\ &= 39^\circ 22.5' \\ &= 39^\circ 22' 30'' \\[6pt] \end{aligned} \] \((ii)\quad -4\) \[ \begin{aligned} \quad -4 &= \dfrac{180}{\pi} \times (-4) \\ &= \dfrac{180 \times 7}{22} \times (-4) \\ &= -\dfrac{2520}{11} \\ &= -229.0909^\circ \\ &= -229^\circ + 0.0909 \times 60' \\ &= -229^\circ 5.454' \\ &= -229^\circ 5' 27'' \\[6pt] \end{aligned} \] \((iii)\quad \dfrac{5\pi}{3}\) \[ \begin{aligned} \quad \dfrac{5\pi}{3} &= \dfrac{180}{\pi} \times \dfrac{5\pi}{3} \\ &= \dfrac{180 \times 5}{3} \\ &= 300^\circ \\[6pt] \end{aligned} \] \((iv)\quad \dfrac{7\pi}{6}\) \[ \begin{aligned} \quad \dfrac{7\pi}{6} &= \dfrac{180}{\pi} \times \dfrac{7\pi}{6} \\ &= 30 \times 7 \\ &= 210^\circ \end{aligned} \]
Hence, the required degree measures are \(39^\circ 22' 30''\), \(-229^\circ 5' 27''\), \(300^\circ\), and \(210^\circ\) respectively.
Q3. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
Solution
Given that the wheel makes 360 revolutions in one minute, the time of one minute is equal to 60 seconds. Hence, the number of revolutions made by the wheel in one second is obtained by dividing 360 by 60.
Since one complete revolution corresponds to \(2\pi\) radians, the total angle turned by the wheel in one second can be calculated as follows.
\[ \begin{aligned} \text{Revolutions per second} &= \dfrac{360}{60} \\ &= 6 \\[6pt] \text{Radians turned in one second} &= 6 \times 2\pi \\ &= 12\pi \end{aligned} \]
Therefore, the wheel turns through \(12\pi\) radians in one second.
Q4. 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 \(\pi=\dfrac{22}{7}\)
Solution
The radius of the circle is 100 cm and the length of the arc is 22 cm. The angle subtended at the centre by the entire circumference of the circle is \(360^\circ\). Hence, the angle subtended at the centre by an arc of length 22 cm can be found using the ratio of the arc length to the circumference.
\[ \begin{aligned} \theta &= \dfrac{360}{2\pi r} \times 22 \\ &= \dfrac{360}{2 \times \dfrac{22}{7} \times 100} \times 22 \\ &= \dfrac{360 \times 7 \times 22}{2 \times 22 \times 100} \\ &= \dfrac{36 \times 7}{20} \\ &= \dfrac{63}{5} \\ &= 12.6^\circ \\ &= 12^\circ + 0.6 \times 60' \\ &= 12^\circ 36' \end{aligned} \]
Therefore, the degree measure of the angle subtended at the centre of the circle is \(12^\circ 36'\).
Q5. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.
Solution
Given that the diameter of the circle is 40 cm, its radius is 20 cm. The length of the given chord is 20 cm.
Let the chord subtend an angle \(2\theta\) at the centre of the circle. From the right triangle formed by drawing a perpendicular from the centre to the chord, half of the chord is 10 cm and the radius is 20 cm.
\[ \begin{aligned} \sin \theta &= \dfrac{10}{20} \\ &= \dfrac{1}{2} \\ \theta &= 30^\circ \\ 2\theta &= 60^\circ \end{aligned} \]
Thus, the angle subtended by the chord at the centre is \(60^\circ\).
The length \(l\) of the minor arc corresponding to this angle is given by
\[ \begin{aligned} l &= \dfrac{2\pi r}{360} \times 60 \\ &= \dfrac{2\pi \times 20}{360} \times 60 \\ &= \dfrac{20\pi}{3} \end{aligned} \]
Hence, the length of the minor arc of the chord is \(\dfrac{20\pi}{3}\) cm.
Q6. If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.
Solution
In the two given circles, arcs of equal length subtend angles \(60^\circ\) and \(75^\circ\) respectively at the centres.
Let the radii of the two circles be \(r_1\) and \(r_2\), and the corresponding angles at the centre be \(\theta_1\) and \(\theta_2\).
\[ \begin{aligned} l_1 &= r_1 \theta_1 \\ l_2 &= r_2 \theta_2 \\ l_1 &= l_2 \\ r_1 \theta_1 &= r_2 \theta_2 \\ \dfrac{r_1}{r_2} &= \dfrac{\theta_2}{\theta_1} \\ &= \dfrac{75}{60} \\ \dfrac{r_1}{r_2} &= \dfrac{5}{4} \\ r_1 : r_2 &= 5 : 4 \end{aligned} \]
Hence, the ratio of the radii of the two circles is \(5 : 4\).
Q7. Find the angle in radian through which a pendulum swings if its length is 75 cm
and th e tip describes an arc of length
(i) 10 cm
(ii) 15 cm
(iii) 21 cm
Solution
The length of the pendulum is 75 cm. When the pendulum swings, the tip describes an arc of a circle whose radius is equal to the length of the pendulum.
\(\left(i\right)\) \[ \begin{aligned} l &= 10 \text{ cm} \\ l &= r\theta \\ \theta &= \dfrac{l}{r} \\ &= \dfrac{10}{75} \\ &= \dfrac{2}{15} \end{aligned} \]
Thus, the angle through which the pendulum swings in this case is \(\dfrac{2}{15}\) radian.
\(\left(ii\right)\) \[ \begin{aligned} \theta &= \dfrac{15}{75} \\ &= \dfrac{1}{5} \end{aligned} \]
Hence, the angle through which the pendulum swings is \(\dfrac{1}{5}\) radian.
\(\left(iii\right)\) \[ \begin{aligned} \theta &= \dfrac{21}{75} \\ &= \dfrac{7}{25} \end{aligned} \]
Therefore, the angle through which the pendulum swings in this case is \(\dfrac{7}{25}\) radian.
