Tag: types of angle

Questions Related to types of angle

Say True or False.
If $m\angle A=53^o$ and $m\angle B=35^o$, then $m\angle A > m\angle B$.

  1. True

  2. False


Correct Option: A
Explanation:
$True$
$m \angle A=53^ \circ  \ and \ m \angle B=35^ \circ$
$\because 53 \ > \ 35$
$\therefore m \angle A> m \angle B$

An angle which measures $0^{\circ}$ is called

  1. Obtuse

  2. Straight

  3. Zero

  4. Right


Correct Option: C
Explanation:

An angle which measures $ {0}^{o} $ is called zero angle.

A, B, C and D are four angles at a point so that $A+B+C+D=4$ rightangles, outof these A and B are acute angles while C and D are obtuse angles. Which of the following relations may be true?

  1. $A+B=C+D$
  2. $A+C=B+D$
  3. $A+D=B+C$

  1. 2 and 3 only

  2. 1 and 3 only

  3. 1 and 2 only

  4. 3 only


Correct Option: A

An angle which measures $\displaystyle 0^{0}$ is called-

  1. Zero

  2. Obtuse

  3. Right

  4. None of these


Correct Option: A
Explanation:

an angle which measures 0 degree is called a zero angle..

In a $\displaystyle \Delta PQR$ PQ = PR and $\displaystyle \angle Q$ is twice that of $\displaystyle \angle P$ Then $\displaystyle \angle Q$__

  1. 75$\displaystyle ^{\circ}$

  2. 65$\displaystyle ^{\circ}$

  3. 72$\displaystyle ^{\circ}$

  4. 100$\displaystyle ^{\circ}$


Correct Option: C
Explanation:

$\displaystyle \because $ PQ = PR 
$\displaystyle \Rightarrow $ $\displaystyle \angle Q=\angle R$
Given that $\displaystyle \angle Q=2\angle P$
We have
$\displaystyle \angle P+\angle Q+\angle R=180^{\circ}$
$\displaystyle \frac{\angle Q}{2}+\angle Q+\angle Q=180^{\circ}$
$\displaystyle \frac{5}{2}\angle Q=180^{\circ}$
$\displaystyle \angle Q=72^{\circ}$        

Find the angle between the lines 3x + 2y = 6 and x + y = 6

  1. 12$\displaystyle ^{\circ}$ 20'

  2. 11$\displaystyle ^{\circ}$ 19'

  3. 14$\displaystyle ^{\circ}$ 25'

  4. 13$\displaystyle ^{\circ}$ 06'


Correct Option: B
Explanation:

Line  $3x+2y=6$

$=>2y=-3x+6$
$=>y=\frac{-3}{2}x+\frac{6}{2}$
$=>m'=\frac{-3}{2}x$
$\therefore tan\theta _1=\frac{-3}{2}$
$=>\theta _1=tan^{-1}(\frac{-3}{2})$
$=>\theta _1=-56.31$
Line is $x+y=6$
$=>y=x+6$
$=>m _2=-1$
$=>tan\theta _2=-1$
$=>\theta _2=tan^{-1}(-1)$
$=>\theta _2=-45$
Thus, ale between them $=(-45-56.31)$
                                          $=11.31^0$

Find the complement of each of the following angles $24^{\circ}$

  1. $66^{\circ}$

  2. $156^{\circ}$

  3. $36^{\circ}$

  4. None of these


Correct Option: A
Explanation:

We know that the complement angle

$=90^0-\theta$

So,
The complement angle of $24^0$ will be
$=90^0-24^0=66^0$ 

Hence, this is the answer.

Find the complement of each of the following angles
$63^{\circ}$

  1. $27^{\circ}$

  2. $54^{\circ}$

  3. $117^{\circ}$

  4. None of these


Correct Option: A
Explanation:

We know that the complement angle

$=90^0-\theta$

So,
The complement angle of $63^0$ will be
$=90^0-63^0=27^0$ 

Hence, this is the answer.

Find the angles in each of the following.
The angle whose complement is one sixth of its supplement

  1. $72^{\circ}$

  2. $32^{\circ}$

  3. $62^{\circ}$

  4. None of these


Correct Option: A
Explanation:

Let the angle be $x^0$.


Supplement angle $=(180^0-x)$

Complement angle $=(90^0-x)$

According to the question,
$\dfrac{1}{6}\times (180^0-x)=(90^0-x)$

$(180^0-x)=6(90^0-x)$

$180^0-x=540^0-6x$

$5x=360^0$

$x=72^0$


Hence, this is the answer.

Find the angles in each of the following.
The angle which is four times its supplement

  1. $144^{\circ}$

  2. $44^{\circ}$

  3. $14^{\circ}$

  4. None of these


Correct Option: A
Explanation:

Let the angle be $x^0$.


According to the question,
Supplement angle $=(180^0-x)\times 4$

So,

$x=720^0-4x$

$5x=720^0$

$x=144^0$


Hence, this is the answer.