Tag: mathematics and statistics

Quizzes (38) Questions (955)

Questions Related to mathematics and statistics

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

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

  1. $0$
  2. $\dfrac{1}{2}$
  3. $\dfrac{3}{4}$
  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

  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 } } $
Reveal answer Fill a bubble to check yourself
B,C Correct answer
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 $\left| \cos { A } -\cos { B }  \right| =2\sin { \cfrac { A+B }{ 2 }  } \sin { \cfrac { B-A }{ 2 }  }$ 
$=2\sin { {30}^{o} } \sqrt { 1-\cos ^{ 2 }{ \cfrac { A-B }{ 2 }  }  } =1\sqrt { 1-\cfrac { 1 }{ 3 }  } =\sqrt { \cfrac { 2 }{ 3 }  }$

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

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

  1. $20^\circ $
  2. $60^\circ $
  3. $\dfrac{1}{3}\,radian$
  4. $\,3\,radian$
Reveal answer Fill a bubble to check yourself
C Correct answer
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,

$\Rightarrow 1=3\times\theta$

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

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

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

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

  4. None
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $4 \sqrt{3}/3$
Multiple choice mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

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

  1. $\pi$
  2. $0$
  3. $\tan^{-1}2$
  4. $3\pi$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
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 $x=\tan ^{ -1 }{ 2 } ,0$

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

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

  1. $152\ cm^{2}$
  2. $\dfrac{154}{3}\ cm^{2}$
  3. $\dfrac{128}{3}\ cm^{2}$
  4. $128\ cm^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Area$=\cfrac { 120 }{ 360 } \times \pi { r }^{ 2 }$
$=\cfrac { \pi  }{ 3 } \times 7\times 7=49\times \cfrac { \pi  }{ 3 } $
$=49\times \cfrac { 22 }{ 7\times 3 } =\cfrac { 154 }{ 3 }cm^2$
Multiple choice mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

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

  1. $\displaystyle 18^{\circ}$
  2. $\displaystyle 36^{\circ}$
  3. $\displaystyle 54^{\circ}$
  4. $\displaystyle 72^{\circ}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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}  $

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

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

  1. $\displaystyle \frac{3\pi ^{c}}{4}$
  2. $\displaystyle \frac{2\pi ^{c}}{3}$
  3. $\displaystyle \frac{4\pi ^{c}}{5}$
  4. $\displaystyle \frac{5\pi ^{c}}{6}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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