Tag: trigonometry

Quizzes (11) Questions (86)

Questions Related to trigonometry

Multiple choice trigonometric equations trigonometric functions trigonometry maths

If $\alpha$ is the angle of first quadrant such that $co\sec ^{ 4 }{ \alpha  }=17+\cot ^{ 4 }{ \alpha  } $, then what is the value of $\sin{\alpha}$?

  1. $\cfrac{1}{3}$
  2. $\cfrac{1}{4}$
  3. $\cfrac{1}{9}$
  4. $\cfrac{1}{16}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$ cosec^{4}\alpha -cot^{4}\alpha = 17$

 (As $ cosec^{2}\alpha -cot^{2}\alpha=1) $

$ \Rightarrow (cosec^{2}\alpha -cot^{2}\alpha )(cosec^{2}\alpha +cot^{2}\alpha ) = 17 $ 

$ \Rightarrow cosec^{2}\alpha +cot^{2}\alpha = 17...(1) $

$ cosec^{2}\alpha -cot^{2}\alpha = 1...(2) $

then $ (1) + (2) \Rightarrow 2cosec^{2}\alpha = 18 $

$ \Rightarrow sin^{2}\alpha = \dfrac{1}{9}\Rightarrow \boxed{sin\,\alpha = \dfrac{1}{3}} $ $ \left ( \because \alpha \,in\,1st\,quadrant \right ) $ 
Multiple choice trigonometric equations trigonometric functions trigonometry maths

General solution of $\dfrac{1-{tan}^{2}x}{{sec}^{2}x}=\dfrac{1}{2}$ is

  1. $n\pi+\dfrac{\pi}{6},n\in Z$
  2. $n\pi-\dfrac{\pi}{6},n\in Z$
  3. $n\pi\pm\dfrac{\pi}{6},n\in Z$
  4. $2n\pi\pm\dfrac{\pi}{6},n\in Z$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given $\dfrac{1-\tan^2 x}{\text{sec}^2 x}=\dfrac{1}{2}$


$\implies \dfrac{1-\tan^2 x}{1+\tan^2 x}=\dfrac{1}{2}$


$\implies 2-2\tan^2 x=1+\tan^2 x$

$\implies \tan^2 x=\dfrac{1}{3}=\tan^2 \dfrac{\pi}{6}$

$\implies x=n\pi\pm \dfrac{\pi}{6}$

Multiple choice trigonometric equations trigonometric functions trigonometry maths

The cosine of the obtuse angle formed by the medians from the vertices of the acute angles of an isosceles right angled triangle is

  1. $- 2 / 3$
  2. $- 4 / 5$
  3. $- 3 / 5$
  4. $- 3 / 4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In an isosceles right triangle, placing vertices at (0,0), (a,0), and (0,a), the medians from the acute angles are calculated. The cosine of the angle between them is -2/3.

Multiple choice trigonometric equations trigonometric functions trigonometry maths

In an isosceles $\triangle ABC$, if the altitudes intersect on the inscribed circle then cosine of the vertical angle $'A'$ is :

  1. $\cfrac{1}{9}$
  2. $\cfrac{1}{3}$
  3. $\cfrac{2}{3}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If the altitudes of an isosceles triangle intersect on the inscribed circle, the geometry dictates that cos(A) = 1/9.

Multiple choice trigonometric equations trigonometric functions trigonometry maths

If $3sin\alpha =5sin\beta ,\quad then\quad \frac { \tan { \frac { \alpha +\beta }{ 2 } } }{ \tan { \frac { \alpha -\beta }{ 2 } } } $ is equal to

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given $3\sin \alpha=5\sin \beta\implies \dfrac{\sin \alpha}{\sin \beta}=\dfrac{5}{3}$


Applying componendo and dividendo rule


$\implies \dfrac{\sin \alpha+\sin \beta}{\sin \alpha-\sin \beta}=\dfrac{5+3}{5-2}$

$\implies \dfrac{2\sin \dfrac{\alpha+\beta}{2}\cos \dfrac{\alpha-\beta}{2}}{2\sin \dfrac{\alpha-\beta}{2}\cos \dfrac{\alpha+\beta}{2}}=\dfrac{8}{2}$

$\implies \dfrac{\tan \dfrac{\alpha+\beta}{2}}{\tan \dfrac{\alpha-\beta}{2}}=4$

Multiple choice trigonometric equations trigonometric functions trigonometry maths

If $y\tan (A+B+C)=x\tan (A+B-C)=\lambda$, then $\tan 2C=?$

  1. $\dfrac{\lambda(x+y)}{\lambda^2-xy}$
  2. $\dfrac{\lambda(x+y)}{\lambda^2+xy}$
  3. $\dfrac{\lambda(x-y)}{xy-\lambda^2}$
  4. $\dfrac{\lambda (x-y)}{xy+\lambda^2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given $y\tan (A+B+C)=x\tan (A+B-C)=\lambda$

$\implies \tan (A+B+C)=\dfrac{\lambda}{y},\tan (A+B-C)=\dfrac{\lambda}{x}$

$\tan 2 C=\tan ((A+B+C)-(A+B-C))=\dfrac{\tan (A+B+C)-\tan (A+B-C)}{1+\tan (A+B+C)\tan (A+B-C)}$

                                                                                $=\dfrac{\frac{\lambda}{y}-\frac{\lambda}{x}}{1+\frac{\lambda^2}{x y}}$

                                                                                $=\dfrac{\lambda(x-y)}{x y+\lambda^2}$

Multiple choice trigonometric equations trigonometric functions trigonometry maths

If $4^{2\, sin^2x}.16^{tan^2x}.2^{4\, cos^2x} = 256 $ such that $0 < x < \dfrac{\pi}{2}$ then $x$ is equal to ___________.

  1. $\dfrac{\pi}{3}$
  2. $\dfrac{\pi}{4}$
  3. $\dfrac{\pi}{12}$
  4. $\dfrac{\pi}{24}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
${4}^{2{\sin}^{2}{x}}.{16}^{{\tan}^{2}{x}}.{2}^{4{\cos}^{2}{x}}=256$

$\Rightarrow\,{2}^{4{\sin}^{2}{x}}.{2}^{4{\tan}^{2}{x}}.{2}^{4{\cos}^{2}{x}}={2}^{8}$

$\Rightarrow\,{2}^{4{\sin}^{2}{x}+4{\tan}^{2}{x}+4{\cos}^{2}{x}}={2}^{8}$

$\Rightarrow\,4{\sin}^{2}{x}+4{\tan}^{2}{x}+4{\cos}^{2}{x}=8$

$\Rightarrow\,{\sin}^{2}{x}+{\tan}^{2}{x}+{\cos}^{2}{x}=2$

$\Rightarrow\,\left({\sin}^{2}{x}+{\cos}^{2}{x}\right)+{\tan}^{2}{x}=2$

$\Rightarrow\,1+{\tan}^{2}{x}=2$ since $\left({\sin}^{2}{x}+{\cos}^{2}{x}=1\right)$

$\Rightarrow\,{\sec}^{2}{x}=2$ since $1+{\tan}^{2}{x}={\sec}^{2}{x}$

$\Rightarrow\,{\cos}^{2}{x}=\dfrac{1}{2}$

$\Rightarrow\,\cos{x}=\pm\dfrac{1}{\sqrt{2}}$

$\Rightarrow\,\cos{x}=\dfrac{1}{\sqrt{2}}$ since $0<x<\dfrac{\pi}{2}$

$\Rightarrow\,x=\dfrac{\pi}{4}$

Multiple choice using trigonometric tables trigonometric ratios of some specific angles trigonometric identities trigonometry maths

Choose the correct option for the following statement.

The line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer.

  1. The given statement is true

  2. The given statement is false

  3. Incomplete information

  4. None of the above.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

By definition,

The line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer.
Therefore, the given statement is true.