Tag: angle and their measurement

Questions Related to angle and their measurement

If $11 \sin^2 x + 7\cos^2x = 8$ then $x =$______

mathematics and statistics angle and their measurement angles and sides naming the sides in a right angled triangle understanding ratios

  1. $nx \pm \dfrac{\pi}{6},\forall n \in Z$

  2. $nx \pm \dfrac{\pi}{4},\forall n \in Z$

  3. $nx \pm \dfrac{\pi}{3},\forall n \in Z$

  4. $nx \pm \dfrac{\pi}{2},\forall n \in Z$

Show answer
Correct Option: A
Explanation:

Given $11\sin^2 x+7\cos^2 x=8$

$\implies 11\sin^2 x+7-7\sin^2 x=8$
$\implies 4\sin^2 x=1$
$\implies \sin^2 x=\dfrac{1}{4}$
$\implies \sin^2 x=\sin^2 \dfrac{\pi}{6}$
$\implies x=n\pi\pm \dfrac{\pi}{6},\forall Z$

If $\alpha, \beta$ are solution of equation a $cos \theta + b sin\theta = c$ then

mathematics and statistics angle and their measurement angles and sides naming the sides in a right angled triangle understanding ratios

  1. $sin \alpha + sin \beta = \dfrac{a^2-c^2}{b^2-a^2}$

  2. $cos \alpha + cos \beta = \dfrac{2ac}{a^2 + b^2}$

  3. $cos \alpha . cos \beta = \dfrac{c^2-b^2}{a^2 + b^2}$

  4. $\sin \alpha.\sin \beta=\dfrac{a^{2}-c^{2}}{b^{2}-a^{2}}$

Show answer
Correct Option: D
Explanation:
$a\cos\theta+b\sin\theta=c$

$a^{2}\cos^{2}\theta+b^{2}\sin^{2}\theta+2ab\sin^{2}\theta.\cos\theta=c^{2}$

$a^{2}(1-\sin^{2}\theta)+b^{2}\sin^{2}\theta+2ab\sin\theta.\cos\theta=c^{2}$

$a^{2}-a^{2}\sin^{2}\theta+b^{2}\sin^{2}\theta\sin^{2}\theta+2ab\sin\theta\cos\theta=c^{2}$

$\sin^{2}\theta(b^{2}-a^{2})+2ab\sin\theta.\cos\theta+a^{2}-c^{2}=0$

So
$\sin \alpha.\sin \beta=\dfrac{a^{2}-c^{2}}{b^{2}-a^{2}}$

If $\cos x + cosy + \cos \theta = 0$ and $\sin x + \sin y + \sin \theta = 0$, then $\cot\left(\dfrac{x + y}{2}\right)$ 

mathematics and statistics angle and their measurement angles and sides naming the sides in a right angled triangle understanding ratios

  1. $\sin \theta$

  2. $\cos \theta$

  3. $\cot \theta$

  4. $\sin\left(\dfrac{x + y}{2}\right)$

Show answer
Correct Option: C
Explanation:

$\cos x+\cos y+\cos \theta=0$

$\cos x+\cos y=-\cos \theta$
$\cfrac { 2\cos(x+y) }{ 2 } \times \cfrac { \cos(x-y) }{ 2 } =-\cos\theta  \longrightarrow 1$
$\sin x+\sin y+\sin \theta=0$
$\sin x+\sin y=-\sin \theta$
$\cfrac { 2\sin(x+y) }{ 2 } \times \cfrac { \sin(x-y) }{ 2 } =-\sin\theta $
Dividing both,
$\cfrac { \cfrac { 2\cos  (x+y) }{ 2 } \times \cfrac { \cos  (x-y) }{ 2 }  }{ \cfrac { 2\sin  (x+y) }{ 2 } \times \cfrac { \sin  (x-y) }{ 2 }  } =\cfrac { -\cos  \theta  }{ -\sin\theta } $
$\cfrac { \cot(x+y) }{ 2 } =\cot\theta $

If $sin:\theta +cos:\theta =p$ and $:tan:\theta +cot:\theta =q$ then $:q\left(p^2-1\right)=$

mathematics and statistics angle and their measurement angles and sides naming the sides in a right angled triangle understanding ratios

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

  2. $2$

  3. $1$

  4. $3$

Show answer
Correct Option: B
Explanation:

$sin:\theta +cos:\theta =p$ 
Squaring on both side we get
$\Rightarrow 1+sin:2\theta =p^2$    $[\because sin:2\theta =p^2-1]$
$Tan\theta +\frac{1}{Tan\theta :}=q$

$\frac{Tan^2\theta +1}{2Tan\theta :}=\frac{q}{2}$

$cosec:2\theta =\frac{q}{2}$

$\Rightarrow \frac{1}{p^2-1}=\frac{q}{2}$   $\Rightarrow 2=\left(p^2-1\right)q$

If $\tan { \theta  } .\tan { (120-\theta ) } .\tan { (120+\theta ) } =\dfrac { 1 }{ \sqrt { 3 }  }$, then $\theta $

mathematics and statistics angle and their measurement angles and sides naming the sides in a right angled triangle understanding ratios

  1. $\dfrac { n\pi }{ 3 } +\dfrac { \pi }{ 18 } ,n\epsilon Z$

  2. $\dfrac { n\pi }{ 3 } +\cfrac { \pi }{ 12 } ,n\epsilon Z$

  3. $\dfrac { n\pi }{ 12 } +\dfrac { \pi }{ 12 } ,n\epsilon Z$

  4. $\dfrac { n\pi }{ 3 } +\dfrac { \pi }{ 6 } ,n\epsilon Z$

Show answer
Correct Option: A

In a $\triangle ABC$, if $a=26, b=30, \cos C=\dfrac{63}{65}$ then $c=$

mathematics and statistics angle and their measurement angles and sides naming the sides in a right angled triangle understanding ratios

  1. $2$

  2. $4$

  3. $6$

  4. $8$

Show answer
Correct Option: D
Explanation:
Given 
In $\triangle ABC,a=26,b=30,\cos C=\dfrac{63}{65}$
As we know that
$c^2 =a^2+b^2-2 a b\cos C=26^2+30^2-2(26)(30)\bigg(\dfrac{63}{65}\bigg)=64$
So $c=8$

If $f ( x ) = \sin x - \dfrac { x } { 2 }$ is increasing function, then

mathematics and statistics angle and their measurement angles and sides naming the sides in a right angled triangle understanding ratios

  1. $0 < x < \dfrac { \pi } { 3 }$

  2. $- \dfrac { \pi } { 3 } < x < 0$

  3. $- \dfrac { \pi } { 3 } < x < \dfrac{\pi}{3}$

  4. None

Show answer
Correct Option: A
Explanation:
$f(x)=\sin x-\dfrac {x}{2}$
$f'(x)=\cos x-\dfrac {1}{2} > 0$
$\cos x > \dfrac {1}{2}$
so $x\in \left (0, \dfrac {\pi}{3}\right)$

In a $\Delta$ABC, $\dfrac{s}{r _1}+\dfrac{s}{r _2}+\dfrac{s}{r _3}-\dfrac{s}{r}$ (where all the symbols have the usual meanings ) is equal to?

mathematics and statistics angle and their measurement angles and sides naming the sides in a right angled triangle understanding ratios

  1. 0

  2. 1

  3. 2

  4. 4

Show answer
Correct Option: A

In $\Delta ABC$, a, b, c are the lengths of its sides and A, B, C are the angles of triangle ABC. The correct relation is 

mathematics and statistics angle and their measurement angles and sides naming the sides in a right angled triangle understanding ratios

  1. $(b-c)sin(\frac{B-C}{2}) =a cos(\frac{A}{2}) $

  2. $(b-c)cos(\frac{A}{2})= a sin(\frac{B-C}{2}) $

  3. $(b+c)sin(\frac{B+C}{2})=a cos(\frac{A}{2}) $

  4. $(b-c)cos(\frac{A}{2}) = 2a sin(\frac{B+C}{2}) $

Show answer
Correct Option: B

Find the product of $\cos{30}^{0}.\cos{45}^{0}.\cos{60}^{0}$

mathematics and statistics angle and their measurement angles and sides naming the sides in a right angled triangle understanding ratios

  1. $0.30$

  2. $0.60$

  3. $0.90$

  4. $0.80$

Show answer
Correct Option: A
Explanation:
$\cos { 30 } \cos { 45 } \cos { 60 } $
$\cfrac { \sqrt { 3 }  }{ 2 } \times \cfrac { 1 }{ \sqrt { 2 }  } \times \cfrac { 1 }{ 2 } =\cfrac { \sqrt { 3 }  }{ 4\sqrt { 2 }  } $