Tag: angle and their measurement

Questions Related to angle and their measurement

In the 5th century who created the table of chords with increasing 1 degree?

  1. Hipparchus

  2. William Rowan Hamilton

  3. Euclid

  4. Ptolemy


Correct Option: D
Explanation:

Ptolemy used length of chords to define his trigonometric functions

The points of discontinuity of $\tan{x}$ are

  1. $n\pi ,n\in I$

  2. $2n\pi ,n\in I$

  3. $(2n+1)\cfrac { \pi }{ 2 } ,n\in I$

  4. None of the above


Correct Option: C
Explanation:

Let $f(x)=\tan {x}$
The points of discontinuity of $f(x)$ are those points where $\tan {x}$ is infinite. This gives
$\tan { x } =\infty $
ie $\tan { x } =\tan { \cfrac { \pi  }{ 2 }  } $
$x=\left( 2n+1 \right) \cfrac { \pi  }{ 2 } ,n\in I$

What is the meaning of trigonometry in Greek language?

  1. measurement

  2. triangle measure

  3. angle measure

  4. degree measure


Correct Option: B
Explanation:

Trigonometry in the Greek language is triangle measure.
(Trigono - triangle + metron- measure)


So, option B is correct.

Find the name of the person who first produce a table for solving a triangle's length and angles.

  1. William Rowan Hamilton

  2. Hipparchus

  3. Euclid

  4. Issac Newton


Correct Option: B
Explanation:

$\text {Hipparchus}$ gave the first table of chords analogus to modern table of sine values, and used them to solve trigonometric problems

What is the value of $\sqrt {2}\sec 45^{\circ} - \tan 30^{\circ}$?

  1. $\dfrac {(2\sqrt {3} - 1)}{3}$

  2. $\dfrac {(\sqrt {3} - 1)}{\sqrt {3}}$

  3. $\dfrac {(2\sqrt {3} - 1)}{\sqrt {3}}$

  4. $\dfrac {(2\sqrt {3} + 1)}{3}$


Correct Option: C
Explanation:

$\sqrt {2}\sec 45^{\circ} - \tan 30^{\circ} = \sqrt {2}\times \sqrt {2} - \dfrac {1}{\sqrt {3}} = \dfrac {2\sqrt {3} - 1}{\sqrt {3}}$.

In triangle $XYZ$, $XZ=YZ$. If the measure of angle $Z$ has ${a}^{o}$, how many degrees are there in the measure of angle $X$?

  1. $x^o=\dfrac {180^o-2a}{2}$

  2. ${ x }^{ o }=\cfrac { { 180 }^{ o }-{ a }^{ o } }{ 2 } $

  3. $x^o=\dfrac {180^o-3a}{3}$

  4. none of these


Correct Option: B
Explanation:

Given, $XZ=YZ$ which implies angles $ZXY$ and $ZYX$ are equal and let it be $\theta$.
We have $ZXY+ZYX+XZY = 180$ , which implies $\theta+\theta+a=180$
Which implies $\theta =\dfrac { (180-a)}{2}$

If $\tan A = \dfrac {1 - \cos B}{\sin B}$, then the value of $\dfrac {2\tan A}{1 - \tan^{2}A}$ is

  1. $\dfrac {(\tan B)}{2}$

  2. $2\tan B$

  3. $\tan B$

  4. $4\tan B$


Correct Option: C
Explanation:

Given, $\tan A =  \dfrac {1 - \cos B}{\sin B} $


                       $= \dfrac {2\sin^{2}\dfrac {B}{2}}{2\sin \dfrac {B}{2}\cdot \cos \dfrac {B}{2}}$


                       $= \tan \dfrac {B}{2}$

Therefore, $A = \dfrac {B}{2} \Rightarrow 2A = B$

Now $\dfrac {2\tan A}{1 - \tan^{2}A} = \tan 2A = \tan B$

The value of sin $15^0$ is

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

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

  3. $\dfrac{-(\sqrt{3}+1)}{2\sqrt{2}}$

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


Correct Option: D
Explanation:

$\sin15^o$


$=\sin(45^o-30^o)$

$=\sin45^o \ \cos30^o - \cos45^o \ \sin30^o$

$=\dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{2}$

$=\dfrac{\sqrt{3}-1}{2\sqrt{2}}$
Hence answer is D

Find number of solutions to the equation:$[ \sin x + \cos x ] = 3 + [ - \sin x ] + [ - \cos x ]$

  1. 0

  2. 1

  3. 2

  4. Infinite


Correct Option: D
Explanation:
$\sin {x}+\cos x=3-\sin x-\cos x$
$2(\sin x+\cos x)=3$
$\sin x+\cos x=1.5$
so these are infinite value of $x$ for which this equation will satisfy

if $\displaystyle Sin\theta =\frac{3}{5}$ what is the value of $\displaystyle  \left ( \tan \theta +\sec \theta  \right )^{2}$?

  1. $2$

  2. $3$

  3. $4$

  4. $-4$


Correct Option: C
Explanation:

$Sin \theta=\dfrac{P}{H}=\dfrac{3}{5}$

According to the Pythagorean therom
$H^2=B^2+P^2$
$\Rightarrow B=\sqrt{h^2-p^2}$
$\Rightarrow B=\sqrt{5^2-3^2}$
$\Rightarrow B=\sqrt{16}$
$\Rightarrow B=4 cm$
$\therefore tan \theta= \dfrac{P}{B}=\dfrac{3}{4}$
$sec \theta=\dfrac{H}{B}=\dfrac{5}{4}$
$\therefore (tan \theta+sec \theta)^2=(\dfrac{3}{4}+\dfrac{5}{4})^2$
$\Rightarrow (\dfrac{8}{4})^2=(2)^2=4$