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Naming the sides in a right angled triangle - class-XI

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Value of $ \displaystyle \sin 45^{\circ} \cos 45 \left ( \tan 45^{\circ}+\cot 45^{\circ} \right )^{2}   $  is 

  1. $1$

  2. $2$

  3. $3$

  4. $4$


Correct Option: B
Explanation:

$ \displaystyle \sin 45^{\circ} \cos 45 \left ( \tan 45^{\circ}+\cot 45^{\circ} \right )^{2}   $

$=\dfrac{1}{\sqrt2} \times \dfrac{1}{\sqrt2} (1+1)^2 $


$=\dfrac42$

$=2$

Two angles are called adjacent if

  1. they lie in the same plane and have a common vertex

  2. they have a ray in common

  3. the intersection of their interiors is empty

  4. all the above


Correct Option: D
Explanation:

Two angles are called adjacent if
they lie in same plane and have a common vertex,they have a ray in common and the intersection of their interiors is empty.
do (D) all the above is correct

In a cyclic quadrilateral ABCD, $\displaystyle \angle ABC=60^{\circ}$ and if O be the centre of the circle then the measure of $\displaystyle \angle OAC$ is

  1. $\displaystyle 20^{\circ}$

  2. $\displaystyle 30^{\circ}$

  3. $\displaystyle 40^{\circ}$

  4. $\displaystyle 50^{\circ}$


Correct Option: B

Triangle measurement is called as _______.

  1. pythagoras

  2. trigonometry

  3. calculus

  4. area of square


Correct Option: B
Explanation:
Trigonometry is the study of triangle measurement.
Hence, the answer is trigonometry.

In the early 9th century AD, _________ produced accurate sine and cosine tables, and the first table of tangents.

  1. Habash al-Hasib al-Marwazi

  2. Muhammad ibn Jabir al-Harrani al-Battani

  3. Muhammad ibn Musa al-Khwarizmi

  4. Abu al-Wafa al-Buzjani


Correct Option: C
Explanation:

In the early 9th century AD, Muhammad ibn Mūsā al-Khwārizmī produced accurate sine and cosine tables, and the first table of tangents. He was also a pioneer in spherical trigonometry.

Who is the founder of trigonometry?

  1. Euclid

  2. Issac Newton

  3. William Rowan Hamilton

  4. Hipparchus


Correct Option: D
Explanation:

Hipparchus the founder of trigonometry. 

So, option D is correct.

The history of trigonometry goes back to the earliest recorded mathematics in Egypt and _____.

  1. German

  2. Indian

  3. Babylon

  4. Japanese


Correct Option: C
Explanation:

The history of trigonometry goes back to the earliest recorded mathematics in Egypt and Babylon.

So, option C is correct.

Trigonometry is used mainly due to the purpose of time keeping and _____.

  1. space

  2. stars

  3. astronomy

  4. planets


Correct Option: C
Explanation:

Trigonometry is used mainly due to the purpose of time keeping and astronomy.

So, option C is correct.

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

  1. William Rowan Hamilton

  2. Hipparchus

  3. Newton

  4. Bartholomaeus Pitiscus


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.

  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:

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

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

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. $nx \pm \dfrac{\pi}{2},\forall n \in Z$


Correct Option: A
Explanation:

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

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

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. $\sin \alpha.\sin \beta=\dfrac{a^{2}-c^{2}}{b^{2}-a^{2}}$


Correct Option: D
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
$\sin \alpha.\sin \beta=\dfrac{a^{2}-c^{2}}{b^{2}-a^{2}}$

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. $\sin\left(\dfrac{x + y}{2}\right)$


Correct Option: C
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 $

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. $3$


Correct Option: B
Explanation:

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

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

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

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

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. $\dfrac { n\pi }{ 3 } +\dfrac { \pi }{ 6 } ,n\epsilon Z$


Correct Option: A

In a $\triangle ABC$, if $a=26, b=30, \cos C=\dfrac{63}{65}$ then $c=$

  1. $2$

  2. $4$

  3. $6$

  4. $8$


Correct Option: D
Explanation:
Given 
In $\triangle ABC,a=26,b=30,\cos C=\dfrac{63}{65}$
As we know that
$c^2 =a^2+b^2-2 a b\cos C=26^2+30^2-2(26)(30)\bigg(\dfrac{63}{65}\bigg)=64$
So $c=8$

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


Correct Option: A
Explanation:
$f(x)=\sin x-\dfrac {x}{2}$
$f'(x)=\cos x-\dfrac {1}{2} > 0$
$\cos x > \dfrac {1}{2}$
so $x\in \left (0, \dfrac {\pi}{3}\right)$

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


Correct Option: A

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}) $


Correct Option: B

Find the product of $\cos{30}^{0}.\cos{45}^{0}.\cos{60}^{0}$

  1. $0.30$

  2. $0.60$

  3. $0.90$

  4. $0.80$


Correct Option: A
Explanation:
$\cos { 30 } \cos { 45 } \cos { 60 } $
$\cfrac { \sqrt { 3 }  }{ 2 } \times \cfrac { 1 }{ \sqrt { 2 }  } \times \cfrac { 1 }{ 2 } =\cfrac { \sqrt { 3 }  }{ 4\sqrt { 2 }  } $

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

The side opposite to the right angle in a right angled triangle is called

  1. Base

  2. Perpendicualr

  3. Hypotenuse

  4. None of these


Correct Option: C
Explanation:

The side opposite to the right angle in a right angled triangle is called Hypotenuse.

The area of the semicircle drawn on the hypotenuse of a right angled triangle is equal to the difference of the areas of the semicircles drawn on the other two sides of the triangles.

  1. True

  2. False


Correct Option: B

If $E. \ tan(x -
30^{\circ}) = j. \ tan(x+120^{\circ})$, then $\frac{E + J}{E-J} =$

  1. $\ sin 2x$

  2. $2 \ cos 2x$

  3. $\ tan2x$

  4. None of these.


Correct Option: B
Explanation:
Given $E\tan (x-30^{\circ})=J\tan (x+120^{\circ})$

$\implies \dfrac{E}{J}=\dfrac{\tan (x+120^{\circ})}{\tan (x-30^{\circ})}$

Applying compoundo and dividendo rule

$\dfrac{E+J}{E-J}=\dfrac{\tan (x-30^{\circ})+\tan (x+120^{\circ})}{\tan (x+120^{\circ})-\tan (x-30^{\circ})}=\dfrac{\frac{\sin (x-30^{\circ})}{\cos (x-30^{\circ})}+\frac{\sin (x+120^{\circ})}{\cos (x+120^{\circ})}}{\frac{\sin (x+120^{\circ})}{\cos (x+120^{\circ})}-\frac{\sin (x-30^{\circ})}{\cos (x-30^{\circ})}}$

                                                        $=\dfrac{\sin (x-30^{\circ})\cos(x+120^{\circ})+\sin (x+120^{\circ})\cos(x-30^{\circ})}{\sin (x+120^{\circ})\cos(x-30^{\circ})-\sin (x-30^{\circ})\cos (x+120^{\circ})}$

                                                       $=\dfrac{\sin (x-30^{\circ}+x+120^{\circ})}{\sin (x+120^{\circ}-x+30^{\circ})}$

                                                      $=\dfrac{\sin (90^{\circ}+2 x)}{\sin 150^{\circ}}$

                                                      $=2\cos 2 x$

A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height 5 meters. At point on the plane, the angle of elevation of the bottom and top of the flag staff are respectively 30$^{\circ}$ and 60$^{\circ}$. The height of tower is

  1. 2m

  2. 5m

  3. 2.5m

  4. 3m


Correct Option: C
Explanation:

Let AB be the tower of height h meter and BC be the height of flag staff surmounted on the tower. Let the point of the plane be D at a distance m meter from the foot of the tower.

In $\bigtriangleup$ ABD,

$\displaystyle tan 30^{\circ} = \frac{AB}{BD}$

$\displaystyle \Rightarrow \frac{1}{\sqrt3} = \frac{h}{x} \Rightarrow x = \sqrt 3 h$.......(1)

In $\displaystyle \bigtriangleup ADC, tan 60^{\circ} = \frac{AC}{AD}$

$\displaystyle \Rightarrow \sqrt3 = \frac{5+h}{x} \Rightarrow x = \frac{5 +h}{\sqrt3}$ .....(2)

From (1) and (2), $\displaystyle \sqrt 3h = \frac{5+h}{\sqrt 3}$

$\Rightarrow 3h = 5 + h \Rightarrow 2h = 5$

$\Rightarrow \displaystyle h = \frac{5}{2} = 2.5 m$

So, the height of tower = 2.5m

If the angle of elevation of a cloud from a point 200 meter above a lake is $\displaystyle 30^{\circ}$ and the angle of depression of its reflection in the lake is $\displaystyle 60^{\circ}$ then the height of the cloud (in meters )above the lake is 

  1. $200$

  2. $300$

  3. $500$

  4. $None$


Correct Option: D
Explanation:

$\tan { 60°=\sqrt { 3 }  } =\dfrac { p }{ b } $

$\tan { 30°=\dfrac { \sqrt { 3 }  }{ 3 }  } =\dfrac { p }{ b } $
$b=200\sqrt { 3 } $
$\tan { 60°=\sqrt { 3 }  } =\dfrac { p }{ 200 } $
$p=200\sqrt { 3 } $
height of the cloud (in meters )above the lake = $200+200\sqrt { 3 } $
Answer none 

If the distance between a 13-foot ladder and a vertical wall is $5$ feet along the ground, how high can a person climb if the ladder is inclined against wall?

  1. $18$ feet

  2. $65$ feet

  3. $\cfrac{13}{5}$ feet

  4. $8$ feet

  5. $12$ feet


Correct Option: E
Explanation:

Length of Ladder = 13 ft

Length between Ladder and Wall = 5 ft
by pythagoras theorum
(hypotenuse)^2 = (perpendicular)^2 + (base)^2
(13)^2 = (height of wall)^2+(5)^2
by solving
Height of Wall = 12 ft

If $sin\theta = 3sin(\theta +2\alpha)$, then the value of $tan(\theta+\alpha)+ 2tan\alpha$ is

  1. 3

  2. 2

  3. 1

  4. 0


Correct Option: D
Explanation:
$\sin \theta =3\sin \left( \theta +2\alpha\right)$

$\Rightarrow 3\sin \left( \theta +\alpha +\alpha \right)=\sin \theta$

      $3\sin \left(\theta+\alpha\right)\cos \alpha +3\cos \left(\theta+\alpha\right)\sin \alpha =\sin \left(\theta+\alpha-\alpha \right)$

      $3\sin \left(\theta+\alpha\right)\cos\alpha +3\cos \left(\theta+\alpha\right)\sin \alpha  =\sin \left(\theta+\alpha\right) \cos \alpha -\sin \alpha \cos \left(\theta+\alpha\right)$

      $2\sin\left(\theta+\alpha\right)\cos \alpha =-4\cos \left(\theta+\alpha\right)\sin \alpha$

      $2\tan \left(\theta+\alpha\right)=-4\tan \alpha$

      $\tan \left(\theta+\alpha\right)=-2\tan \alpha$

$\Rightarrow \tan \left(\theta+\alpha\right)+2\tan \alpha=0$

Hence, the answer is $0.$

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