Tag: angle and their measurement

Quizzes (6) Questions (67)

Questions Related to angle and their measurement

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

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

  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. $nx \pm \dfrac{\pi}{2},\forall n \in Z$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

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

  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. $\sin \alpha.\sin \beta=\dfrac{a^{2}-c^{2}}{b^{2}-a^{2}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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
$\sin \alpha.\sin \beta=\dfrac{a^{2}-c^{2}}{b^{2}-a^{2}}$
Multiple choice mathematics and statistics angle and their measurement angles and sides naming the sides in a right angled triangle understanding ratios

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

  1. $\sin \theta$
  2. $\cos \theta$
  3. $\cot \theta$
  4. $\sin\left(\dfrac{x + y}{2}\right)$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $

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

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

  1. $\frac{1}{2}$
  2. $2$
  3. $1$
  4. $3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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

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

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

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

  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. $\dfrac { n\pi }{ 3 } +\dfrac { \pi }{ 6 } ,n\epsilon Z$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the identity tan(theta) * tan(60-theta) * tan(60+theta) = tan(3*theta), we have tan(3*theta) = 1/sqrt(3). Thus, 3*theta = n*pi + pi/6, which simplifies to theta = n*pi/3 + pi/18.

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

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

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

  4. None
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$f(x)=\sin x-\dfrac {x}{2}$
$f'(x)=\cos x-\dfrac {1}{2} > 0$
$\cos x > \dfrac {1}{2}$
so $x\in \left (0, \dfrac {\pi}{3}\right)$
Multiple choice mathematics and statistics angle and their measurement angles and sides naming the sides in a right angled triangle understanding ratios

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?

  1. 0

  2. 1

  3. 2

  4. 4
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the formulas r1 = s*tan(A/2), r2 = s*tan(B/2), r3 = s*tan(C/2), and r = s*tan(A/2)*tan(B/2)*tan(C/2), the expression simplifies to 0 based on trigonometric identities in a triangle.

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

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 

  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}) $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the Law of Sines and Mollweide's formulas, the correct relation is (b-c)cos(A/2) = a*sin((B-C)/2).