Tag: angle and their measurement

Questions Related to angle and their measurement

Trigonometry is a branch of mathematics that studies relationships involving lengths and ______ of triangles.

  1. radian

  2. degree

  3. angle

  4. vector


Correct Option: C
Explanation:

Trigonometry is a branch of mathematics that studies relationships involving lengths and angles of triangles.

So, option C is correct.

The term trigonometry was first invented by the German mathematician ______.

  1. William Rowan Hamilton

  2. Euclid

  3. Newton

  4. Bartholomaeus Pitiscus


Correct Option: D
Explanation:

The term trigonometry was first invented by the German mathematician Bartholomaeus Pitiscus.

So, option D is correct.

______ mathematicians created the trigonometry system based on the sine function instead of the chords.

  1. Greek

  2. Indian

  3. German

  4. Egyptian


Correct Option: B
Explanation:
Indian mathematicians $\text{Aryabhata}$ created the trigonometry system based on the sine function instead of the chords.
.
Hence, the answer is Indian.

Who published the trigonometry in 1595?

  1. William Rowan Hamilton

  2. Hipparchus

  3. Bartholomaeus Pitiscus

  4. Newton


Correct Option: C
Explanation:

Trigonometry was first published by Bartholomaeus Pitiscus in 1595.

So, option C is correct.

In $\Delta ABC$ if $a=8,b=9,c=10$, then the value of $\dfrac{{\tan C}}{{\sin B}}$ is

  1. $\dfrac{{32}}{9}$

  2. $\dfrac{{24}}{7}$

  3. $\dfrac{{21}}{4}$

  4. $\dfrac{{18}}{5}$


Correct Option: A
Explanation:

In $\triangle ABC,a=8,b=9,c=10$

as we know that
$\dfrac{sin C}{\sin B}=\dfrac{c}{b}=\dfrac{10}{9}$
and also $\cos C=\dfrac{a^2+b^2-c^2}{2 a b}=\dfrac{8^2+9^2-10^2}{2\times 8\times 9}=\dfrac{5}{16}$
$\dfrac{\tan C}{\sin B}=\dfrac{\sin C}{\sin B}\times \dfrac{1}{\cos C}=\dfrac{10}{9}\times \dfrac{16}{5}=\dfrac{32}{9}$

If $\sin \theta + \cos \theta = 1$, then what is the value of $\sin \theta \cos \theta$?

  1. $2$

  2. $0$

  3. $1$

  4. $\dfrac {1}{2}$


Correct Option: B
Explanation:

Given, $\sin \theta + \cos \theta = 1$

Squaring both sides gives,
$\sin^{2}\theta + \cos^{2}\theta + 2\sin \theta \cos \theta = 1$
$\Rightarrow 1 + 2\sin \theta \cos \theta = 1$
$\Rightarrow 2\sin \theta \cos \theta = 0$
$\Rightarrow \sin \theta \cos \theta = 0$

If $t _1=(\tan x)^{\cot x}, t _2=(\cot x)^{\cot x}, t _3=(\tan x)^{\tan x}, t _4=(\cot x)^{\tan x}, 0 < x < \dfrac{\pi}{4}$, then:

  1. $t _1 < t _2 < t _3 < t _4$

  2. $t _2 > t _4 > t _3 > t _1$

  3. $t _1 > t _4 > t _3 > t _2$

  4. $t _1 > t _2 > t _3 > t _4$


Correct Option: B
Explanation:

Since $0 < x < \dfrac{\pi}{4}$

$\therefore \tan x < 1$ and $\cot x > 1$ .....$(1)$

$\therefore$ Choose $\tan x=1-k _1$ and $\cot x=1+k _2$, where 

$k _1$ and $k _2$ are very small $+$ve quantities.

$\therefore t _3=(1-k _1)^{1-k _1}, t _1=(1-k _1)^{1+k _2}$  ....$(2)$

$t _4=(1+k _2)^{1-k _1}, t _2=(1+k _2)^{1+k _2}$  .....$(3)$

$\therefore t _4 > t _3$ by $(3)$ and $(2)$, $t _2 > t _4$ by $(2)$ and $(3)$

$\therefore t _2 > t _4 > t _3$. Also $t _3 > t _1$

$\therefore t _2 > t _4 > t _3 > t _1$.

The angle of elevation and angle of depression both are measured with

  1. the vertical only

  2. the horizontal only

  3. both horizontal and vertical

  4. NONE OF THE ABOVE


Correct Option: B
Explanation:

The angle of elevation and angle of depression are measured with Line of Sight which is always Horizontal.

For a
positive integer n,
let
${f _n}\left( \theta  \right) = \left( {\tan \frac{\theta }{2}} \right)\left( {1 + \sec \theta } \right)\left( {1 + \sec 2\theta } \right)\left( {1 + \sec {2^2}\theta } \right)...\left( {1 + \sec {2^n}\theta } \right),then$

  1. ${f _2}\left( {\frac{\pi }{{16}}} \right) = 1$

  2. ${f _3}\left( {\frac{\pi }{{32}}} \right) = 1$

  3. ${f _4}\left( {\frac{\pi }{{64}}} \right) = 1$

  4. ${f _5}\left( {\frac{\pi }{{128}}} \right) = 1$


Correct Option: A,B,C,D
Explanation:

$f _n(\theta)=(\tan \frac{\theta}{2})(1+\text{sec}\theta)(1+\text{sec} 2\theta)(1+\text{sec}2^2 \theta)\cdots(1+\text{sec}2^n \theta)$

           $=\dfrac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}\times \dfrac{1+\cos \theta}{\cos \theta}\times \dfrac{1+\cos 2\theta}{\cos 2\theta}\times \dfrac{1+\cos 2^2 \theta}{\cos 2^2 \theta}\cdots\times \dfrac{1+\cos 2^n \theta}{\cos 2^n \theta}$
           $=\dfrac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}\times \dfrac{2\cos ^2 \frac{\theta}{2}}{\cos \theta}\times \dfrac{2\cos ^2 \theta}{\cos 2 \theta}\times \dfrac{2\cos^2 2\theta}{\cos 2^2 \theta}\times \cdots\times \dfrac{2\cos 2^{n-1}\theta}{\cos^2 2^n\theta}$
           $=(2\sin \frac{\theta}{2}\cos \frac{\theta}{2})\times (2\cos \theta)\times (2\cos 2 \theta)\times \cdots\times \dfrac{2\cos 2^{n-1}\theta}{\cos 2^n \theta}$
           $=(2\sin \theta\cos \theta)\times (2\cos 2 \theta)\times \cdots\times \dfrac{2\cos 2^{n-1}\theta}{\cos 2^n \theta}$
           $=\dfrac{\sin 2^n \theta}{\cos 2^n \theta}=\tan 2^n \theta$

$f _2\bigg(\dfrac{\pi}{16}\bigg)=\tan (2^2\times \dfrac{\pi}{16})=\tan \frac{\pi}{4}=1$
$f _3\bigg(\dfrac{\pi }{32}\bigg)=\tan (2^3\times \dfrac{\pi}{32})=\tan \frac{\pi}{4}=1$
$f _4\bigg(\dfrac{\pi}{64}\bigg)=\tan (2^4\times \dfrac{\pi}{64})=\tan \frac{\pi}{4}=1$
$f _5\bigg(\dfrac{\pi}{128}\bigg)=\tan (2^5\times \dfrac{\pi}{128})=\tan \frac{\pi}{4}=1$

$8\sin { \theta  } \cos { \theta  } .\cos { 2\theta  } \cos { 4\theta  } =\sin { x } \Longrightarrow x=$?

  1. x=-8θx=8θ

  2. $x=8\theta$

  3. x=4θx=8θ

  4. None of these


Correct Option: B
Explanation:

$8\sin\theta \cos\theta\cos 2\theta \cos 4\theta =\sin x$

$=4\sin 2 \theta\cos 2\theta\cos 4 \theta$
$=2\sin 4\theta\cos 4\theta$
$\sin 8\theta=x$
$x=8\theta$