\(\angle QPR=75^{ 0 }\)
\(\angle PQR+\angle QPR=105^{ 0 }\)

n=5
2n-4=2x5-4=6

\(\angle BC||DE\) and \(\angle CDE=90^{ 0 }\quad \)
\(\therefore \angle BCD=90^{ 0 }\) | Alternate Interior angles
\(\angle CDE=90^{ 0 }\) 
\(\therefore \angle DCE=180^{ 0 }-(30^{ 0 }+90^{ 0 })=60^{ 0 }\)

The sum of consecutive interior angles on the same side of a transversal is \(180^{ 0 }\)

\(y=62^{ 0 }\)
\(x+62^{ 0 }=180^{ 0 }\Rightarrow x=118^{ 0 }\) 
\(\therefore \angle x-\angle y=118^{ 0 }-62^{ 0 }=56^{ 0 }\)

\(40^{ 0 }+3y+2y=180^{ 0 }\)

Linear Pair Axiom

Required angle=\(180^{ 0 }\)-\(60^{ 0 }\)=\(120^{ 0 }\)

Definition of complementary angles

Given,OT 丄 PR, ㄥTQR = 40° and ㄥSPR= 30°,
In ΔTQR, ㄥTQR + ㄥQTR + ㄥTRQ = 180° [by angle sum property of a triangle]
⇒ 40° + 90° + x = 180° ⇒ 130° + x = 180°
⇒ x =180°-130°  ⇒ x = 500
Now, ㄥPSQ is an exterior angle for ΔPSR.
ஃ  ㄥPSQ = ㄥSPR + ㄥSRP  [by theorem 2]
⇒ y = 30° + x  ⇒  y = 30° + 50° [ஃ x = 50°]
⇒ y = 80°
Hence, x = 50° and y = 80°

\(75^{ 0 },285^{ 0 }\)

\(20^{ 0 }\)

Given AB||CD, CD||EF, \(EA\bot AB\)
\(\angle BEF=55^o\)
y = 180^o- 55^o = 125^o(Co-interior angles)
x = y = 125^o(Corresponding angles)
\(EA\bot AB\)
\(EA\bot AB\)
Z + 55^o = 90^o
Z = 90^o - 55^o
= 35^o

Let \(\angle AOB=2x\) and \(\angle BOC=3x\)
\(\angle AOB+\angle BOC=\angle AOC\)
2x+3x=75^o
\(\Rightarrow x=\frac{75^o}{5}=15^o\)
\(\angle AOB=2x=30^o\) and \(\angle BOC=3x=45^o\)

(i) Let 'O' be the point from where the three friend move to the points A, Band C, such that OA makes equal angles with OB and OC such that
∠AOB= ∠AOC ...(i)
Also, OB makes equal angles with OA and OC

Such that
∠AOB= ∠BOC ...(ii)
From (i) and (ii), we get
∠AOB = ∠BOC = ∠AOC ...(iii)
Since, sum of all angles around a point = 360^о
⇒ ∠AOB + ∠BOC + ∠AOC = 360^о
⇒ ∠AOB + ∠AOB + ∠AOB = 360^o [from iii]
⇒ 3 ∠AOB = 360^о
⇒ ∠AOB = \(\frac{360^{\circ}}{3}\) = 120^o
∴ From (iii), we have:
∠BOC = 120^o and ∠AOC = 120^o
(ii) Lines and Angles (Geometry)
(iii) Helping attitude towards disastrous victims.

Let the angles of the given triangle measure (2x )^o, (3x)^o and (5x)^o.
∵ Sum of the angles of a triangle is 180°.
∴ (2x)^о + (3x)^o + (5x)^o = 180^o
⇒ 10x = 180°
or x = \(\frac{180^{\circ}}{10}\) = 18°
∴ 2x = 2 x 18 = 36°
3x = 3 x 18 = 54°
5x = 5 x 18 = 90°
∴ The measures of the angles of the given triangle are 36°, 54° and 90°.

Let us consider ΔABC and through A draw DE ॥ BC.

∵ BC ॥ DE and AB is a transversal.
∠4 = ∠1 [Interior alternate angles] ... (1)
Again, BC ‖ DF and AC is a transversal.
∴ ∠5 = ∠2 [Interior alternate angles] ... (2)
Adding (1) and (2), we have
∠4 + ∠5 = ∠1 + ∠2
Adding ∠3 on both sides, we have
∠4 + ∠5 + ∠3 = ∠1 + ∠2 + ∠3
Since, DAF is a straight line,
∴ ∠4 + ∠3 + ∠5 = 180°
∠1 + ∠2 + ∠3 = 180°
i.e. ∠ABC + ∠BCA + ∠BAC = 180°
∴ The sum of angles of a triangle is 180°.

∵ \(\left.\begin{array}{l} \mathrm{AB} \| \mathrm{CD} \\ \mathrm{EF} \| \mathrm{CD} \end{array}\right\}\) (given)

∴ AB II EF and PQ is a transversal.
∴ Interior alternate angles are equal.
∴  ∠x = ∠y ...(1)
Again, AB II CD,
∴ Interior opposite angles are supplementary,
⇒ y + z = 180^o
But y : z = 3 : 7 
∴  y = \(\frac{180^{\circ}}{(y+z)}\) x 3 = \(\left[\frac{180^{\circ}}{(3+7)}\right]\) x 3 = \(\frac{180^{\circ}}{10}\) x 3 = 54^о
and z = \(\frac{180^{\circ}}{(y+z)}\) x 7 = \(\left[\frac{180^{\circ}}{(3+7)}\right]\) x 7 = \(\frac{180^{\circ}}{10}\) x 7 = 126^о ...(2)
From (1) and (2), we have
x = 126^о

Since AB is a straight line,
∴ ∠AOC + ∠COE + ∠EOB = 180^о
or (∠AOC + ∠BOE) + ∠COE = 180^о
or 70^о + ∠COE = 180^о [∴ ∠AOC + ∠BOC = 70^o (Given)]
or ∠COE 180^o - 70^о = 110^o
∴ Reflex ∠COE = 360° - 110° = 250°
∵ AB and CD intersect at O.
∴ ∠COA = ∠BOD [Vertically opposite angles]
But ∠BOD = 40° [Given]
∴ ∠COA = 40°
Also ∠AOC + ∠BOE = 70°
∴ 40° + ∠BOE = 70°
or ∠BOE = 70° - 40° = 30°
Thus, ∠BOE = 30° and reflex ∠COE = 250.

We know that the sum of angles that are formed on a straight line is equal to 180°.
So, Reason is correct
Also, the sum of all the interior angles of a triangle is 1800 Let the angles measure (3x)°, (5x)° and (7x)°.
Then,3x + 5x + 7x = 180° ⇒15x = 180° ⇒x = 12°
Therefore, the angles are 3(12)°=36°, 5(12)°=60° and 7(12)° = 84°. Hence, the triangle is acute-angled.
So, Assertion (A) is also true.
But reason (R) is not the correct explanation of assertion (A). Correct option is (b) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A).

We know that the sum of angles that are formed on a straight line is equal to 180°.
So, Reason is correct
We have : ∠AOC+∠BOC=180°   [Since AOB is a straight line ]
⇒3x + 10 + 4x − 26 = 180°
⇒7x = 196°
⇒x = 28°
∴∠BOC = [4 × 28 − 26]°
⇒∠BOC=86°.
So, Assertion (A) is also true.
Correct option is (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

We know that the sum of all the interior angles of a triangle is 1800. So, Reason (R) is true.
Since the Sum of two obtuse angles will be more than 1800. So, Assertion (A) is false.
Correct option is (d) Assertion (A) is false but reason (R) is true.

We know that two angles are said to be supplementary if their sum of measure of angles is 180^о.
So, Reason is correct.
Let the angle be x . Supplement of x = \(\frac{1}{4}\)x
So, x + \(\frac{1}{4}\)x = 180^о ⇒ \(\frac{5}{4}\)x = 180^о ⇒ x = 144^о
So, Assertion is also correct
Also, reason (R) is the correct explanation of assertion (A).
Correct option is (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

We know that the sum of linear pair of angles is always 1800.
So, Reason is correct.
Now, a + b = 40^о + 150^о = 190^о ≠180^о
Hence, Assertion is not correct
Correct option is (d) Assertion (A) is false but reason (R) is true.