Tag: mathematics and statistics

Questions Related to mathematics and statistics

If $\tan 45^{\circ} = \cot \theta$, then the value of $\theta$, in radians is

mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

  1. $\pi$

  2. $\dfrac{\pi}{9}$

  3. $\dfrac{\pi}{4}$

  4. $\dfrac{\pi}{12}$

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Correct Option: C
Explanation:

$\tan 45^{\circ} = \cot \theta$
$\tan \theta = \cot \theta$   only for $\theta = \pi/ 4$.
Hence, $\theta = 45^{\circ}$ or $\pi/ 4$.

The value of $cos^{2}30^{0}-cos^{2}60^{0}-cos 60^{0}$ is

mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

  1. $0$

  2. $\dfrac{1}{2}$

  3. $\dfrac{3}{4}$

  4. $1$

Show answer
Correct Option: A
Explanation:

$cos^{2}30^{0}-cos^{2}60^{0}-cos 60^{0}={ \left( \frac { \sqrt { 3 }  }{ 2 }  \right)  }^{ 2 }-{ \left( \frac { 1 }{ 2 }  \right)  }^{ 2 }-\frac { 1 }{ 2 } =\frac { 3 }{ 4 } -\frac { 1 }{ 4 } -\frac { 1 }{ 2 } =\frac { 3-1-2 }{ 4 } =0$

If $A+B=\dfrac { \pi  }{ 3 } $ and $\cos { A } +\cos { B } =1 $, then which of the following are true: 

mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

  1. $\cos { \left( A-B \right) =\dfrac { 1 }{ 3 } } $

  2. $\cos { \left( A-B \right) =-\dfrac { 1 }{ 3 } } $

  3. $\left| \cos { A } -\cos { B } \right| =\sqrt { 2/3 } $

  4. $\left| \cos { A } -\cos { B } \right| =\cfrac { 1 }{ \sqrt { 3 } } $

Show answer
Correct Option: B,C
Explanation:

Ans. $(b)$, $(c)$

From the given relation, we have 
$2\cos { \cfrac { A+B }{ 2 }  } \cos { \cfrac { A-B }{ 2 } =1 }$
Or $2\cos { {30}^{o} } \cos { \cfrac { A-B }{ 2 }  } =1$
$\therefore\quad \cos { \cfrac { A-B }{ 2 }  } =\cfrac { 1 }{ \sqrt { 3 }  }$
$\therefore\quad \cos { \left( A-B \right)  } =\cos ^{ 2 }{ \cfrac { A-B }{ 2 } -1 } =2.\cfrac { 1 }{ 3 } -1=-\cfrac { 1 }{ 3 } \Rightarrow \left( b \right)$
Again $\left| \cos { A } -\cos { B }  \right| =2\sin { \cfrac { A+B }{ 2 }  } \sin { \cfrac { B-A }{ 2 }  }$ 
$=2\sin { {30}^{o} } \sqrt { 1-\cos ^{ 2 }{ \cfrac { A-B }{ 2 }  }  } =1\sqrt { 1-\cfrac { 1 }{ 3 }  } =\sqrt { \cfrac { 2 }{ 3 }  }$

The angle subtended at the centre of circle of radius $3$ metres by an arc of length $1$ metre is equal to

mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

  1. $20^\circ $

  2. $60^\circ $

  3. $\dfrac{1}{3}\,radian$

  4. $\,3\,radian$

Show answer
Correct Option: C
Explanation:

We know that 

$l=r\times\theta$

Where $l\rightarrow arc$ $length$
            $r\rightarrow radius$
            $\theta\rightarrow angle$ $subtended$ $by$ $the$ $arc$

Substituting the values of these terms we get,

$\Rightarrow 1=3\times\theta$

$\Rightarrow\theta=\dfrac{1}{3} radian$

The value of $\dfrac{1}{\cos 290^o}+\dfrac{1}{\sqrt{3}\sin 250^o}$ is?

mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

  1. $\dfrac{2\sqrt{3}}{3}$

  2. $\dfrac{4\sqrt{3}}{3}$

  3. $\sqrt{3}$

  4. None

Show answer
Correct Option: B
Explanation:
Here, $\dfrac{1}{ \cos 290^{o}} + \dfrac{1 }{ \sqrt{3} \sin 250^{o}}$ 
$= \dfrac{1}{ \cos (270+20)^{o}} + \dfrac{1}{ \sqrt{3} \sin (270-20)^{o}}$
as we know, $\cos (270+A)= \sin A$
& $\sin (270- B)= - \cos B$
So, $=\dfrac{1}{\sin 20}+ \dfrac{1}{\sqrt{3} (- \cos 20)}$
$=\dfrac{- \sqrt{3} \cos 20+ \sin 20}{- \sqrt{3} \cos 20 \cos 20}$
$=\dfrac{- (\sqrt{3} \cos 20 - \sin 20)}{- \sqrt{3} \sin 20 \cos 20}$
$ =\dfrac{ \sqrt{3} \cos 20- \sin 20}{\sqrt{3} \sin 20 \cos 20}$
(Multiply & Divide in Numerator & denominator by $2$ we get.  )
$=\dfrac{2 \left( \dfrac{\sqrt{3}}{2} \cos 20- \dfrac{1}{2} \sin 20  \right)}{\dfrac{\sqrt{3}}{2} (2 \sin 20 \cos 20)}$
$=\dfrac{2 (\sin 60 \cos 20- \cos 60 \sin 20)}{\dfrac{\sqrt{3}}{2} (\sin 40)}$
$=\dfrac{4}{ \sqrt{3}} \dfrac{\sin (60-20)}{\sin (40)}=\dfrac{4}{\sqrt{3}}= \dfrac{4\sqrt{3}}{3} $
So, value is $4 \sqrt{3}/3$

Find the degree measure corresponding to $\left(\dfrac{1}{6}\right)^C$.

mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

  1. $9.549^\circ$

  2. $9^032$ $43.6"$

  3. $10^0$

  4. None

Show answer
Correct Option: A
Explanation:

$\pi$ rad $=180^o$

$\therefore 1\ rad=\dfrac{180}{\pi}=57.296^o$
$\therefore (1/6)^c=1/6\times \dfrac{180}{\pi}=9.549^o$

If $\cos x=\sqrt{1-\sin2x},0\le x\le \pi$, then possible  value of $x$ is 

mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

  1. $\pi$

  2. $0$

  3. $\tan^{-1}2$

  4. $3\pi$

Show answer
Correct Option: B,C
Explanation:

$\cos { x } =\sqrt { 1-\sin { 2x }  } $; $x\in (0,\pi)$

$=\sqrt { 1-2\sin { x } .\cos { x }  } =\sqrt { \sin ^{ 2 }{ x } -2\sin { x } \cos { x } +\cos ^{ 2 }{ x }  } \left[ \because 1=\sin ^{ 2 }{ x } +\cos ^{ 2 }{ x } ,\forall x\in R \right] $
$=\sqrt { { \left( \sin { x } -\cos { x }  \right)  }^{ 2 } } \left[ \because \sqrt { { x }^{ 2 } } =\left| x \right|  \right] $
$\cos { x } =\left| \sin { x } -\cos { x }  \right| $
case I
$\sin { x } \ge \cos { x } ,x\in \left[ 0,\pi  \right] \Rightarrow \left| \sin { x } -\cos { x }  \right| =\sin { x } -\cos { x } $
$\therefore \log { x } =\sin { x } -\cos { x } $
$\therefore \cos { x } =\sin { x } -\cos { x } \Rightarrow 2\cos { x } =\sin { x } \Leftrightarrow \tan { x } =2\Rightarrow x=\tan ^{ -1 }{ 2 } \left[ \because x\in \left[ 0,\pi  \right]  \right] $
case II
$\sin { x } <\cos { x } ;x\in \left[ 0,\pi  \right] $
$\Rightarrow \left| \sin { x } -\cos { x }  \right| =\cos { x } -\sin { x } $
$\therefore \cos { x } =\cos { x } -\sin { x } \Rightarrow \sin { x } =0\Rightarrow x=0,\pi $
but $x=\pi$ is rejected as $\cos (\pi)=-1$
$\therefore$ only $x=0$
Finally $x=\tan ^{ -1 }{ 2 } ,0$

The area of a sector of a circle of radius $7\ cm$ and central angle $120^{o}$ is 

mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

  1. $152\ cm^{2}$

  2. $\dfrac{154}{3}\ cm^{2}$

  3. $\dfrac{128}{3}\ cm^{2}$

  4. $128\ cm^{2}$

Show answer
Correct Option: B
Explanation:
Area$=\cfrac { 120 }{ 360 } \times \pi { r }^{ 2 }$
$=\cfrac { \pi  }{ 3 } \times 7\times 7=49\times \cfrac { \pi  }{ 3 } $
$=49\times \cfrac { 22 }{ 7\times 3 } =\cfrac { 154 }{ 3 }cm^2$

$\displaystyle \frac{\pi ^{c}}{5}$ in sexagesimal measure is _____

mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

  1. $\displaystyle 18^{\circ}$

  2. $\displaystyle 36^{\circ}$

  3. $\displaystyle 54^{\circ}$

  4. $\displaystyle 72^{\circ}$

Show answer
Correct Option: B
Explanation:

In $\text{Sexagesimal System}$, an angle is measured in degrees, minutes and seconds.
$ \pi = {180}^{0} $

So, $ \dfrac {\pi}{5} = \dfrac {{180}^{0}}{5} = {36}^{0}  $

The value of $\displaystyle 144^{\circ}$ in circular measure is ___ 

mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

  1. $\displaystyle \frac{3\pi ^{c}}{4}$

  2. $\displaystyle \frac{2\pi ^{c}}{3}$

  3. $\displaystyle \frac{4\pi ^{c}}{5}$

  4. $\displaystyle \frac{5\pi ^{c}}{6}$

Show answer
Correct Option: C
Explanation:

$ {144}^{0} = {144}^{0} \times \dfrac {{\pi}^{c}}{{180}^{0}} = \dfrac {4{\pi}^{c}}{5} $