Tag: degree measure of angle

Questions Related to degree measure of angle

$\dfrac{\cos(90-A)\sin(90-A)}{\tan (90-A)}$=

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

  1. $\sin^{2}A$

  2. $\cos^{2}A$

  3. $\sin A$

  4. $1$

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Correct Option: A
Explanation:
$ =  \dfrac{cos\,(90-A)\,sin\,(90-A)}{tan(90-A)}$
$ = \dfrac{sin\,A \times cos\,A}{\dfrac{cos\,A}{sin\,A}} = \dfrac{sin^{2}A\times cos\,A}{cos\,A}$
$ = sin^{2}A$

The value of $ \displaystyle 36^{\circ} $ in radians is 

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

  1. $ \displaystyle \frac{\pi }{2} $

  2. $ \displaystyle \frac{2\pi }{5} $

  3. $ \displaystyle \frac{\pi }{5} $

  4. $ 3\pi $

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Correct Option: C
Explanation:

$\displaystyle 180^{\circ}      $    =  $\displaystyle \pi       $ radians 


$\displaystyle  \therefore  1^{\circ}=\frac{\pi }{180}    radians     $
$\displaystyle  \Rightarrow 36^{\circ}=\frac{\pi }{180}\times 36 \ radians = \frac{\pi }{5}radians      $

A unit radian is approximately equal to

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

  1. $57^{\circ} 17' 43"$

  2. $57^{\circ} 17' 45"$

  3. $57^{\circ} 17' 47"$

  4. $57^{\circ} 17' 49"$

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Correct Option: B
Explanation:
1 radian x 180 degrees per pi radians = 57.295779513082 degrees.

0.295779513082 degrees x 60 minutes per 1 degree = 17.74677078492 minutes.
0.74677078492 minutes x 60 seconds per 1 minute = 44.8062470952 seconds.

Answer : 57 degrees, 17.75 minutes or

57 degrees, 17 minutes, 45 seconds

The value of $\cot 15^{\circ} \cot 20^{\circ} \cot 70^{\circ} \cot 75^{\circ}$ is equal to

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

  1. $-1$

  2. $0$

  3. $1$

  4. $2$

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Correct Option: C
Explanation:

The value of $\cot 15^{\circ}.\cot 20^{\circ}.\cot 70^{\circ}.\cot 75^{\circ}$ is
$= (\cot 15^{\circ} . \cot 75^{\circ}) (\cot 20^{\circ} \cot 70^{\circ})$

Now $\cot 75^{\circ} = \tan 15^{\circ}$ and $\cot 70^{\circ} = \tan 20^{\circ}$   ....................... $\cot(90-\theta)=\tan\theta$

Therefore, the given expression can be written as 
$ (\cot 15^{\circ} \tan 15^{\circ})(\cot 20^{\circ} \tan 20^{\circ})$
$= 1\times 1 = 1$

Consider the following statements :
1. $1^o$ in radian measure is less than 0.02 radians.
2. 1 radian in degree measure is greater than $45^o$ 
Which of the above statements is/are correct ?

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

  1. 1 only

  2. 2 only

  3. Both 1 and 2

  4. Neither 1 nor 2

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Correct Option: C
Explanation:

1 degree in radians $=1\times \dfrac { \pi  }{ 180 } =0.0174$  which is less than $0.02$ radians.


1 radians in degree $=1\times \dfrac { 180 }{ \pi  } =57.32$ which is greater than $45^{o}$.
Hence, both are correct.

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. $1$

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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 } } $

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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 $4 \sqrt{3}/3$