Interior and exterior of an angle - class-IX

Given,
$\angle P - \angle Q = 42$ (I)
$\angle Q - \angle R = 21$ (II)
Sum of angles = 180
$\angle P + \angle Q + \angle R = 180$ (III)
Add all the three equations:
$2 \angle P + \angle Q = 180 + 42 + 21$
$2 \angle P + \angle Q = 243$ (IV)

Add I and IV,
$2 \angle P + \angle P = 243 + 42$
$3 \angle P = 285$
$\angle P = 95^{\circ}$
From IV, $\angle Q = 243 - 2\times 95$
$\angle Q = 53^{\circ}$
From II,
$\angle R = 53 - 21 = 32^{\circ}$

Answer is option B (False)
In any triangle, if one of the sides is produced, then the exterior angle is greater than either of the interior and opposite angles.
Hence the above statement is false

Angle remains the same if we extend the arms.

When an arm of an angle is extended then the measure of angle remains the same.

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

An $angle$ is a combination of two rays (half-lines) with a common endpoint. The latter is known as the vertex of the $angle$ and the rays as the sides, sometimes as the legs and sometimes the arms of the $angle.$

An angle is measured with reference to a circle with its centre at the common endpoint of the rays. Hence, the sum of angles at a point is always 360 degrees.

The sum of exterior angle of a triangle is $ { 360 } $
the interior angle of a triangle add upto  $ { 180 } $, For any given corner exterior angle is $ { 180 } $ minus interior angle.
So sum of exterior angle is (180-a)+(180-b)+(180-c)=3*180-(a+b+c)=540-180=$360^0$

$\displaystyle \because $ Sum of the angles in a triangle = $ \displaystyle  180^{\circ} $
i.e. $ \displaystyle 75^{\circ}+95^{\circ}+x=180^{\circ} $
$ \displaystyle \Rightarrow  x=10^{\circ} $

Two rays of angles are called two sides of the angle..

Complete angle = $\displaystyle 360^{0}$

The common end point of an angle is called a vertex..

As the sum of angles of a triangle is 180 degree therefore first 3 options are ruled out & hence 4th is the answer.

The angle formed by the page of an open is obtuse .

Trisecting an angle means dividing it in three equal parts.

Angle at a point completes one rotation, i.e. $360^{\circ}$

Since, O is ethocentric,
$\angle BOC +\angle A = 180$
$3 \angle A = 180$
$\angle A = 60$

Therefore, $\angle BOC = 120^{\circ}$

The sum of exterior angles of any convex polygon is 360 degrees. For a regular polygon, each exterior angle is 360/n. Here, 360/n = 40, so n = 360/40 = 9.

$1$

A triangle has 3 sides, 3 vertices, at these each vertices, we get 2 exterior angles,

An angle of measure $0^o$ is called a zero degree angle.

An angle of measure $90^o$ is called a right angle.

An angle of measure $360^o$ is called a complete angle. [ Using definition of complete angle]

Angles of the triangle are in the ratio $3 : 4: 5$
Let the angles be $3x, 4x$ and $5x$
Sum of the angles of the triangle $= 180$
Thus, $3x + 4x + 5x = 180$
$12x = 180$
$x = 15$
Thus, the angles are $45^{o}, 60^{o}$ and $75^{o}$.

Given,

The two rays of an angle $=$ 2 sides of the angle 

The maximum number of letters that can be used to represent an angle are $3$

The two rays of an angle are called two sides of the angle

We know, The interior angles of a triangle always add up to 180.
So,  statement A, B and C is not correct.
(C) An exterior angle of a triangle is always greater than the opposite interior angles.
 is correct
Answer (C) An exterior angle of a triangle is always greater than the opposite interior angles.

required angle = $\frac{360}{48}$
$\frac{30}{4}$ = $\frac{15}{2}$= $\left ( 7\frac{1}{2} \right )^0$

Let, required angle is =x

Supplementary angle of x is =${{180}^{0}}-x$

Now, according to question,

$ 5\times x={{180}^{0}}-x $

$ 6x={{180}^{0}} $

$ x={{30}^{0}} $

Hence, this is the answer.

Here, the angle is written as $\angle{ROP}$ and the angle is made by the intersection of two lines. 

The common end point where the two rays meet is called as a vertex.

To draw an angle of 150° using a pair of compass and 

Obtuse angles are those angles whose measure is more than$90°$ but less than $180°$.

2 $\angle 1 + \angle B = 180^{circ}$         (linear pair)
$\angle 1 = 90^{\circ}-\dfrac{1}{2}\angle B$          (1)
Similarly,
$\angle 2 = 90^{\circ} - \dfrac{1}{2} \angle C$   (2)
$\angle BOC = 180^{circ}-  (\angle 1 + \angle 2)$
=$180^{circ}- [180^{circ}-\dfrac{1}{2}$ ($\angle B + \angle C$)]
=$\dfrac{1}{2}[\angle B + \angle C] = \dfrac{1}{2} (180^{circ} - \angle A) = 90^{circ}-\dfrac{1}{2} \angle A$

A half turn is a rotation of 180 degrees. In either clockwise or counter-clockwise direction, this is represented as +180 or -180 degrees.