Tag: maths

Questions Related to maths

Find sec $\theta$, if $\theta$ be an angle in standard position with (x, y) a point on the terminal side of $\theta$ and r = $\sqrt{x^{2}+y^{2}}\neq 0$.

  1. $\dfrac{y}{r}$

  2. $\dfrac{r}{y}$

  3. $\dfrac{x}{r}$

  4. $\dfrac{r}{x}$


Correct Option: D
Explanation:

By using Pythagoras theorem, we will find the value of hypotenuse.
$x^{2}+y^{2}=r^{2}$

So, sec $\theta$ = $\frac{hypotenuse}{adjacent\space\ side }$= $\dfrac{r}{x}$

sec $\theta$ = $\dfrac{r}{x}$

So, option D is correct.

Find the exact value of sin $\theta$ trigonometric functions for the angle formed when the terminal side passes through (6, 8).

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

  2. $\dfrac{6}{5}$

  3. $\dfrac{10}{5}$

  4. $\dfrac{3}{5}$


Correct Option: A
Explanation:

By using Pythagoras theorem, we will find the value of hypotenuse.
$6^{2}+8^{2}=c^{2}$
$36 + 64 = c^{2}$
$c = 10$
So, $\sin \theta$ = $\dfrac{opposite \space\ side }{hypotenuse}$

= $\dfrac{8}{10}$ = $\dfrac{4}{5}$

So, option A is correct.

Find the exact value of cosec $\theta$ trigonometric functions for the angle formed when the terminal side passes through $(3, 4).$

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

  2. $\dfrac{3}{5}$

  3. $\dfrac{5}{3}$

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


Correct Option: C
Explanation:

By using Pythagoras theorem, we will find the value of hypotenuse.
$3^{2}+4^{2}=c^{2}$
$9 + 16 = c^{2}$
$c = 5$
So, $\csc \theta$ = $\dfrac{hypotenuse}{opposite \space\ side}$

$\csc \theta$ = $\dfrac{5}{3}$

So, option C is correct.

Find cot $\theta$, if $\theta$ be an angle in standard position with (x, y) a point on the terminal side of $\theta$ and r = $\sqrt{x^{2}+y^{2}}\neq 0$.

  1. $\dfrac{y}{r}$

  2. $\dfrac{x}{y}$

  3. $\dfrac{x}{r}$

  4. $\dfrac{r}{x}$


Correct Option: B
Explanation:

By using Pythagoras theorem, we will find the value of hypotenuse.
$x^{2}+y^{2}=r^{2}$
So, cot $\theta$ = $\dfrac{adjacent\space\ side}{opposite\space\ side}$= $\dfrac{x}{y}$


cot $\theta$ = $\dfrac{x}{y}$

So, option B is correct.

Find cosec $\theta$, if $\theta$ be an angle in standard position with (x, y) a point on the terminal side of $\theta$ and r = $\sqrt{x^{2}+y^{2}}\neq 0$.

  1. $\dfrac{y}{r}$

  2. $\dfrac{r}{y}$

  3. $\dfrac{x}{r}$

  4. $\dfrac{r}{x}$


Correct Option: B
Explanation:

By using Pythagoras theorem, we will find the value of hypotenuse.
$x^{2}+y^{2}=r^{2}$
So, cosec $\theta$ = $\dfrac{hypotenuse}{opposite\space\ side }$= $\dfrac{r}{y}$

cosec $\theta$ = $\dfrac{r}{y}$

So, option B is correct.


lf the distance between the points $(a \cos\theta, a \sin\theta), (a \cos\phi, a \mathrm{sin} \phi)$ is $2\mathrm{a}$ then $\theta=$.

  1. $2n\pi\pm\pi+\phi, n\in z$

  2. $n\displaystyle \pi\pm\frac{\pi}{2}+\phi, n\in z$

  3. $n\pi-\phi, n\in z$

  4. $2n\pi+\phi, n\in z$


Correct Option: A
Explanation:
Given points $(a=cos\theta, a\sin\theta), (a\cos\phi, a\sin \phi)$

distance $2a$

distance between $(x _{1}, y _{1}), (x _{2}, y _{2})$

$\sqrt{(x _{2}-x _{1})^{2}+(y _{2}-y _{2})^{2}}$

So, 
$\sqrt{(a\cos\theta-a\cos\theta)^{2}+(a\sin\phi-a\sin\theta)^{2}}=2a$

Squaring on both sides

$\Rightarrow a^{2}(\cos^{2}\theta+\cos^{2}\phi-2\cos\theta\cos\phi)+a(\sin^{2}\phi+\sin^{2}\theta-2\sin\theta-\sin\phi)=4a^{2}$

$\Rightarrow a^{2}(\sin^{2}\theta+\cos^{2}\theta+\sin^{2}\theta+\cos^{2}\theta-2(\sin\theta-\sin\phi+\cos\theta\cos\phi))=4a^{2}$

$\Rightarrow a^{2}(1+1-2(\cos(\theta-\phi))=4a^{2}$

$\Rightarrow 2-2\cos(\theta-\phi)=4$

$\Rightarrow 2\cos(\theta-\phi)=-2$

$\Rightarrow \cos(\theta-\phi)=-1=\cos(\pi)$

$\Rightarrow \theta-\phi=2n\pi\pm \pi, n\in z$

$\Rightarrow \boxed{\theta=2n\pi\pm\pi+\phi, n\in z}$

Find the exact value of $\dfrac{\sin \theta}{\sec \theta}$ trigonometric functions for the angle formed when the terminal side passes through $(3, 4)$.

  1. $\dfrac{25}{12}$

  2. $\dfrac{12}{5}$

  3. $\dfrac{12}{25}$

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


Correct Option: C
Explanation:

By using Pythagoras theorem, we will find the value of hypotenuse.
$3^{2}+4^{2}=c^{2}$
$9 + 16 = c^{2}$
$c = 5$

So, $\dfrac{\sin\theta}{\sec\theta}$ = $\dfrac{opposite \space\ side \times adjacent \space side}{hypotenuse \times hypotenuse}$

= $\dfrac{3\times 4}{5\times 5}$

=$\dfrac{12}{25}$

So, option C is correct.

A rectangle of length $4\ cm$ and breadth $3\ cm$ is scaled up $2$ times. What is the new length of the rectangle?

  1. $6\ cm$

  2. $4\ cm$

  3. $8\ cm$

  4. $3\ cm$


Correct Option: C
Explanation:

The size of the rectangle become double when it is scaled up $2$ times.

So new length $=2\times 4=8\ \ cm$
So option $C$ is correct.

A circle of radius $7\ cm$ is scaled $3$ times. Then the perimeter of the circle become:

  1. $3$ times the original perimeter

  2. $6$ times the original perimeter

  3. $9$ times the original perimeter

  4. Doesn't change


Correct Option: A
Explanation:

Radius of circle $=7 \ \ cm$

Perimeter $=2\pi r=2\times \pi\times7=14\pi\ \ cm$
When scaled $3$ times
New radius $=3\times 7=21 \ \ cm$
New Perimeter $=2\pi r=2\times \pi\times21=42\pi\ \ cm$
Ratio of perimeters $=\dfrac{42\pi}{14\pi}=3$
So the perimeter becomes three times.

If one shape becomes another using a resize, then the shapes are __________. 

  1. similar

  2. congruent

  3. mirror images

  4. none of the above


Correct Option: A
Explanation:

Resizing leads to change in scale factor and if the scale factor remains equal, then the figures will be similar to each other.