SolveItClass 11 · NCERT

NCERT Solutions · Class 11 Mathematics Trigonometric Functions

52 questions · 52 still being checked

EXERCISE 3.1 1–7 (part 1 of 6)

  1. Exercise 1

    Find the radian measures corresponding to the following degree measures:
    (i)
    25\displaystyle 25°
    (ii)
    -47\displaystyle 47°30\displaystyle 30'
    (iii)
    240\displaystyle 240°
    (iv)
    520\displaystyle 520°

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    NCERT’s answer
    (i)
    $\displaystyle \frac{5 \pi}{36}$ (ii) $\displaystyle -\frac{19 \pi}{72}$ (iii) $\displaystyle \frac{4 \pi}{3}$ (iv) $\displaystyle \frac{26 \pi}{9}$
    Degree to radian. Since \(\displaystyle 180^{\circ}=\pi\) radian, \(\displaystyle 1^{\circ}=\dfrac{\pi}{180}\) radian, so multiply each degree measure by \(\displaystyle \dfrac{\pi}{180}\).
    (i)
    \[25^{\circ}=25\times\frac{\pi}{180}=\frac{5\pi}{36}\ \text{radian}\]
    (ii)
    Clear the minutes first: \(\displaystyle 30'=\left(\dfrac{30}{60}\right)^{\circ}=\left(\dfrac{1}{2}\right)^{\circ}\), so \(\displaystyle -47^{\circ}30'=-\left(\dfrac{95}{2}\right)^{\circ}\).
    \[-\frac{95}{2}\times\frac{\pi}{180}=-\frac{95\pi}{360}=-\frac{19\pi}{72}\ \text{radian}\]
    (iii)
    \[240^{\circ}=240\times\frac{\pi}{180}=\frac{4\pi}{3}\ \text{radian}\]
    (iv)
    \[520^{\circ}=520\times\frac{\pi}{180}=\frac{26\pi}{9}\ \text{radian}\]
    (i) \(\displaystyle \dfrac{5\pi}{36}\) (ii) \(\displaystyle -\dfrac{19\pi}{72}\) (iii) \(\displaystyle \dfrac{4\pi}{3}\) (iv) \(\displaystyle \dfrac{26\pi}{9}\) radian.

  2. Exercise 2

    Find the degree measures corresponding to the following radian measures (Use π=227\displaystyle \pi=\frac{22}{7} ).
    (i)
    1116\displaystyle \frac{11}{16}
    (ii)
    4\displaystyle 4
    (iii)
    5π3\displaystyle \frac{5 \pi}{3}
    (iv)
    7π6\displaystyle \frac{7 \pi}{6}

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    NCERT’s answer
    (i)
    $\displaystyle 39$° $\displaystyle 22$' $\displaystyle 30$″ (ii) -$\displaystyle 229$° $\displaystyle 5$' $\displaystyle 27$'' (iii) $\displaystyle 300$° (iv) $\displaystyle 210$°
    Radian to degree. Since \(\displaystyle \pi\) radian \(\displaystyle =180^{\circ}\), one radian \(\displaystyle =\left(\dfrac{180}{\pi}\right)^{\circ}\). With \(\displaystyle \pi=\dfrac{22}{7}\),
    \[\frac{180}{\pi}=180\times\frac{7}{22}=\frac{630}{11}\]
    so multiply each radian measure by \(\displaystyle \dfrac{630}{11}\). (In (iii) and (iv) the \(\displaystyle \pi\) cancels, so no approximation is needed there.)
    (i)
    \[\frac{11}{16}\times\frac{630}{11}=\frac{630}{16}=39.375^{\circ}\]
    Now \(\displaystyle 0.375^{\circ}=0.375\times60'=22.5'=22'30''\), so the measure is \(\displaystyle 39^{\circ}22'30''\).
    (ii)
    \[-4\times\frac{630}{11}=-\frac{2520}{11}=-229\tfrac{1}{11}\ \text{degrees}\]
    Here \(\displaystyle \dfrac{1}{11}^{\circ}=\dfrac{60}{11}'=5\tfrac{5}{11}'\) and \(\displaystyle \dfrac{5}{11}'=\dfrac{300}{11}''\approx 27''\), giving \(\displaystyle -229^{\circ}5'27''\) (nearest second).
    (iii)
    \[\frac{5\pi}{3}\ \text{radian}=\frac{5\pi}{3}\times\frac{180}{\pi}=5\times60=300^{\circ}\]
    (iv)
    \[\frac{7\pi}{6}\ \text{radian}=\frac{7\pi}{6}\times\frac{180}{\pi}=7\times30=210^{\circ}\]
    (i) \(\displaystyle 39^{\circ}22'30''\) (ii) \(\displaystyle -229^{\circ}5'27''\) (approx.) (iii) \(\displaystyle 300^{\circ}\) (iv) \(\displaystyle 210^{\circ}\).

  3. Exercise 3

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

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    NCERT’s answer
    $\displaystyle 12 \pi$
    Revolutions to radians. One complete revolution is an angle of \(\displaystyle 2\pi\) radian.In one minute the wheel makes \(\displaystyle 360\) revolutions, so in one second it makes \[\frac{360}{60}=6\ \text{revolutions}.\] Hence the angle turned in one second is \[6\times 2\pi=12\pi\ \text{radian}.\]The wheel turns through \(\displaystyle 12\pi\) radian (\(\displaystyle \approx 37.7\) radian) in one second.

  4. Exercise 4

    Find the degree measure of the angle subtended at the centre of a circle of radius 100\displaystyle 100 cm by an arc of length 22\displaystyle 22 cm (Use π=227\displaystyle \pi=\frac{22}{7} ).

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    NCERT’s answer
    $\displaystyle 12^{\circ} 36^{\prime}$
    Arc-length formula. For an arc of length \(\displaystyle l\) on a circle of radius \(\displaystyle r\), the angle subtended at the centre is \(\displaystyle \theta=\dfrac{l}{r}\) radian.Here \(\displaystyle l=22\) cm and \(\displaystyle r=100\) cm, so \[\theta=\frac{22}{100}=\frac{11}{50}\ \text{radian}.\]Convert to degrees using \(\displaystyle 1\) radian \(\displaystyle =\left(\dfrac{180}{\pi}\right)^{\circ}=180\times\dfrac{7}{22}=\dfrac{630}{11}\) degrees: \[\theta=\frac{22}{100}\times\frac{630}{11}=\frac{630}{50}=12.6^{\circ}.\] Since \(\displaystyle 0.6^{\circ}=0.6\times 60'=36'\), this is \(\displaystyle 12^{\circ}36'\).The angle subtended is \(\displaystyle 12.6^{\circ}=12^{\circ}36'\).

  5. Exercise 5

    In a circle of diameter 40\displaystyle 40 cm, the length of a chord is 20\displaystyle 20 cm. Find the length of minor arc of the chord.

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    NCERT’s answer
    $\displaystyle \frac{20 \pi}{3}$
    Chord equal to the radius. The diameter is \(\displaystyle 40\) cm, so the radius is \(\displaystyle r=20\) cm, and the chord \(\displaystyle AB=20\) cm.Let \(\displaystyle O\) be the centre and \(\displaystyle AB\) the chord; a sketch of the circle with the two radii \(\displaystyle OA\), \(\displaystyle OB\) drawn to the ends of the chord makes the next step obvious. In triangle \(\displaystyle OAB\), \(\displaystyle OA=OB=20\) cm (radii) and \(\displaystyle AB=20\) cm, so the triangle is equilateral. Therefore the central angle is \[\angle AOB=60^{\circ}=\frac{\pi}{3}\ \text{radian}.\]The minor arc cut off by the chord is the arc standing on this angle, so by \(\displaystyle l=r\theta\), \[l=20\times\frac{\pi}{3}=\frac{20\pi}{3}\ \text{cm}\approx 20.94\ \text{cm}.\]Length of the minor arc \(\displaystyle =\dfrac{20\pi}{3}\) cm.

  6. Exercise 6

    If in two circles, arcs of the same length subtend angles 60\displaystyle 60° and 75\displaystyle 75° at the centre, find the ratio of their radii.

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    NCERT’s answer
    $\displaystyle 5$ : $\displaystyle 4$
    Equal arcs, \(\displaystyle l=r\theta\). Let the radii be \(\displaystyle r_{1}\) and \(\displaystyle r_{2}\), subtending \[\theta_{1}=60^{\circ}=\frac{\pi}{3}\ \text{radian},\qquad \theta_{2}=75^{\circ}=\frac{5\pi}{12}\ \text{radian}.\] (The formula \(\displaystyle l=r\theta\) is valid only when \(\displaystyle \theta\) is in radian, which is why the degrees are converted first.)The two arcs have the same length \(\displaystyle l\), so \[r_{1}\theta_{1}=r_{2}\theta_{2}\quad\Longrightarrow\quad \frac{r_{1}}{r_{2}}=\frac{\theta_{2}}{\theta_{1}}=\frac{5\pi/12}{\pi/3}=\frac{5\pi}{12}\times\frac{3}{\pi}=\frac{5}{4}.\]The radii are in the ratio \(\displaystyle 5:4\).

  7. Exercise 7

    Find the angle in radian through which a pendulum swings if its length is 75\displaystyle 75 cm and th e tip describes an arc of length
    (i)
    10\displaystyle 10 cm
    (ii)
    15\displaystyle 15 cm
    (iii)
    21\displaystyle 21 cm

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    NCERT’s answer
    (i)
    $\displaystyle \frac{2}{15}$ (ii) $\displaystyle \frac{1}{5}$ (iii) $\displaystyle \frac{7}{25}$
    Arc-length formula. The tip of the pendulum moves on a circle of radius \(\displaystyle r=75\) cm, and \(\displaystyle \theta=\dfrac{l}{r}\) radian.
    (i)
    \[\theta=\frac{10}{75}=\frac{2}{15}\ \text{radian}\]
    (ii)
    \[\theta=\frac{15}{75}=\frac{1}{5}\ \text{radian}\]
    (iii)
    \[\theta=\frac{21}{75}=\frac{7}{25}\ \text{radian}\]
    (i) \(\displaystyle \dfrac{2}{15}\) (ii) \(\displaystyle \dfrac{1}{5}\) (iii) \(\displaystyle \dfrac{7}{25}\) radian.