Tag: mathematics and statistics

Questions Related to mathematics and statistics

The first recorded use of trigonometry came from the Hellenistic mathematician ________________.

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

  1. William Rowan Hamilton

  2. Hipparchus

  3. Newton

  4. Bartholomaeus Pitiscus

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Correct Option: B
Explanation:

The first recorded use of trigonometry came from the Hellenistic mathematician Hipparchus.

So, option B is correct.

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

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

  1. radian

  2. degree

  3. angle

  4. vector

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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 ______.

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

  1. William Rowan Hamilton

  2. Euclid

  3. Newton

  4. Bartholomaeus Pitiscus

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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.

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

  1. Greek

  2. Indian

  3. German

  4. Egyptian

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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?

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

  1. William Rowan Hamilton

  2. Hipparchus

  3. Bartholomaeus Pitiscus

  4. Newton

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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

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

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

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

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

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

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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 $\sin \theta \cos \theta$?

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

  1. $2$

  2. $0$

  3. $1$

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

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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:

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

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

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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

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

  1. the vertical only

  2. the horizontal only

  3. both horizontal and vertical

  4. NONE OF THE ABOVE

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

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

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

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

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

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

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