Tag: maths

Questions Related to maths

$\displaystyle3\tan^2{30^\circ}+\frac{4}{3}\cos^2{30^\circ}-2\sin^2{45^\circ}-\frac{1}{3}\sin^2{60^\circ}$ is equal to__________________.

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

  1. $\displaystyle\frac{1}{4}$

  2. $\displaystyle\frac{3}{4}$

  3. $1$

  4. $0$

Show answer
Correct Option: B
Explanation:

Given 


$3\tan ^230+\dfrac 43\cos ^230-2\sin ^245-\dfrac 13\sin ^260$

$=3\left(\dfrac 1{\sqrt 3}\right)^2+\dfrac 43\left(\dfrac {\sqrt 3}2 \right)^2-2\left(\dfrac 1{\sqrt 2}\right)^2-\dfrac 13\left(\dfrac {\sqrt 3}2\right)^2$

$=1+1-1-\dfrac 14$

$= \dfrac 34$

The angle measuring $\displaystyle \frac{\pi ^{c}}{4}$ when expressed in centesimal system is ___ 

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

  1. $\displaystyle 50^{g}$

  2. $\displaystyle 60^{g}$

  3. $\displaystyle 75^{g}$

  4. $\displaystyle 100^{g}$

Show answer
Correct Option: A
Explanation:

In $\text{Centesimal System}$, an angle is measured in grades, minutes and seconds.
Given angle $ = \dfrac {{\pi}^c}{4} = \dfrac {{180}^{0}}{4} = {45}^{0} $

We know that $ {1}^{0} = {(\dfrac {10}{9})}^{g} $


$ \Rightarrow {45}^{0} = {\dfrac {10}{9}} \times 45^{g}  ={50}^{g} $

$\displaystyle 30^{\circ}$ in centesimal measure is _____

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

  1. $\displaystyle \frac{50^{g}}{3}$

  2. $\displaystyle \frac{100^{g}}{3}$

  3. $\displaystyle \frac{160^{g}}{3}$

  4. $\displaystyle \frac{200^{g}}{3}$

Show answer
Correct Option: B
Explanation:

In $\text{Centesimal System}$, an angle is measured in grades, minutes and seconds.
In centesimal system, we know that $ {1}^{0} = {(\dfrac {10}{9})}^{g} $
$ \Rightarrow  {30}^{0} =  {(\dfrac {10}{9})}^{g} \times 30 = {(\dfrac {100}{3})}^{g} $

When the sun is $30^o$ above the horizon, what is the length of the shadow cast by a building $40$ m high?

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

  1. $50.23$ m

  2. $70.24$ m

  3. $68.25$ m

  4. $69.28$ m

Show answer
Correct Option: D
Explanation:

$\tan 30^o = \dfrac{40m}{shadow}$

Shadow $= \dfrac{40}{\tan 30^o}$
$= \dfrac{40}{\frac{1}{\sqrt{3}}}$= $\dfrac{40}{0.57735}$
The shadow of the building is $69.28$ m.

So, option D is correct.

From the tower $30$ m above the sea, the angle of depression of a boat is $68^o$. How far is the boat from the tower?

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

  1. $12.12$ m

  2. $11.11$ m

  3. $10.10$ m

  4. $9.99$ m

Show answer
Correct Option: A
Explanation:
$\tan 68^o = \dfrac{30}{distance}$
Distance $=\dfrac{30}{\tan 68^o}$
Distance = $\dfrac{30}{2.475087}$
Distance $= 12.12$ m

So, option A is correct.

When the sun is $50^o$ above the horizon, how long is the shadow cast by a building $16$ m high?

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

  1. $23$ m

  2. $13.42$ m

  3. $43.42$ m

  4. $23.42$ m

Show answer
Correct Option: B
Explanation:

$\tan 50^o = \dfrac{16m}{shadow}$

Shadow $= \dfrac{16}{\tan 50^o}$= $\dfrac{16}{1.191754}$= $13.42$
The shadow of the building is $13.42$ m

So, option B is correct.

The equation of a line parallel to $x+2y=1$ and passing through the point of intersection of the lines $x-y=4$ and $3x+y=7$ is ?

graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

  1. $x+2y=5$

  2. $4x+8y-1=0$

  3. $4x+8y+1=0$

  4. none of these

Show answer
Correct Option: B
Explanation:

Let the equation of line parallel to $x+2y=1$ be $y=mx+c$

intersection point of line $x-y=4$ and $3x+y=7$ is given by solving two equation we get $x=\dfrac { 11 }{ 4 } $ and $y=\dfrac { -5 }{ 4 } $
from equation $x+2y=1\ \Rightarrow y=-\dfrac { 1 }{ 2 } x+\dfrac { 1 }{ 2 } $
we get ${ m } _{ 1 }=-\dfrac { 1 }{ 2 } $
since $y=mx+c$ is parallel to $x+2y=1$
$\therefore { m } _{ 2 }=-\dfrac { 1 }{ 2 } $
also, $y=mx+c\ \Rightarrow \dfrac { -5 }{ 4 } =\dfrac { -1 }{ 2 } .\dfrac { 11 }{ 4 } +c\ \Rightarrow c=\dfrac { 1 }{ 8 } $
therefore equation of line parallel to $x+2y=1$ is given as 
$y=mx+c\ \Rightarrow y=\dfrac { -1 }{ 2 } x+\dfrac { 1 }{ 8 } \ \Rightarrow 4x+8y-1=0$

$2x + y = 0$ is the equation of a diameter of the circle which touches the lines $4x-3y+10=0$ and $4x-3y-30=0$ The center and radius of the circle are ?

graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

  1. $\left (-2, 1\right) ; 4$

  2. $\left (1, -2\right) ; 8$

  3. $\left (1, -2\right) ; 4$

  4. $\left (1, -2\right) ; 16$

Show answer
Correct Option: C
Explanation:
Given $4x-3y+10=0$ and $4x-3y-30=0$ touches circle implies they are tangent.

Solving the line $2x+y=0$  and  $4x-2y+10=0 $
$x=-1$ and $y=2 $ Point A

Solving the line $2x+y=0$  and  $4x-3y-30=0$
$x=3; y=-6 $ Point B

Distance between the parallel lines is length of diameter
$d=\dfrac{(C _1-C _2)}{\sqrt{(a^2+b^2)}}\\$
$d=\dfrac{(10-(-30)}{\sqrt{(16+9)}}\\$
$d=\dfrac{10+30}{5}$
$d=8$
$r=4$

O is midpoint of AB
$(x,y)=\dfrac{3-1}{2}, \dfrac{2-6}{2}$ $=(1,-2)$

The line $y = x$ meets $y = ke^x , k \le 0$ at

graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

  1. no ponits

  2. one point

  3. two points

  4. none of these

Show answer
Correct Option: B

Let a, b, c and d be non-zero numbers. If the point of intersection of the lines $4ax+2ay+c=0$ and $5bx+2by+d=0$ lies in the fourth quadrant and is equidistant from the two axes, then:

graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

  1. $2bc-3ad =0$

  2. $2bc+3ad =0$

  3. $3bc -2ad =0$

  4. $3bc +2ad =0$

Show answer
Correct Option: C
Explanation:

If it lies in the fourth quadrant, we get$(x,-x)$

$2ax+c = 0$ and $3bx+d = 0$
$\cfrac{c}{2a} = \cfrac{d}{3b}$
$3bc-2ad = 0$