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Correct option is (A) 65°

\(\because\) Sum of complementary angle is \(90^\circ.\)

\(\therefore\) Complementary angle to \(25^\circ\)

\(=90^\circ-25^\circ=65^\circ\)

Correct option is  A) 65°

Let an angle be x

Its complement = 90° – x

According to question

x = 90° – x – 25°

⇒ 2x = 65°

⇒ x = 32\(\frac { { 1 }^{ 0 } }{ 2 }\)

Hence, measurement of the required angle be 32\(\frac { { 1 }^{ 0 } }{ 2 }\).

i) The line which intersects two or more lines at distinct points is called transversal. 

ii) If the pairs of alternate interior angles are equal then the lines are parallel. 

iii) The sum of interior angles on the same side of the transversal are supplementary then the lines are parallel. 

iv) If two lines intersect each other then the number of common points they have Only one.

Correct option is (D) 70°

\(\because\) Sum of supplementary angles is \(180^\circ.\)

\(\therefore\) Supplementary angle to \(110^\circ\)

\(=180^\circ-110^\circ=70^\circ\)

Correct option is  D) 70°

Here we are given that POS is a straight line

∠POQ + ∠QOR + ∠ROS = 180° (straight angle)

⇒ 60° + 5x – 20° + 40° = 180°

⇒ 5x + 80° = 180°

⇒ 5x = 100°

⇒ x = 20°

∠QOR = 5x – 20° 

= 5 x 20° – 20° = 80°

∴ l || m

⇒ ∠1 = ∠3 (Vertically opposite angles)

Also ∠1 = ∠5 (Corresponding angles)

∠5 = ∠7 (Vertically opposite angles)

Hence, ∠3, ∠5 and ∠7 are equivalent to ∠1.

∠1 + ∠2 = 100° (given)

But ∠1 = ∠2 (vertically opposite angles)

⇒ 2∠2 = 100°

⇒ ∠2 = 50°

Also ∠2 + ∠4 = 180° (linear pair)

⇒ ∠4 = 180° – 50° = 130°

Since ray OC stands on line AB. 

Therefore

∠AOC + ∠COB = 180° (Linear pair of angles)

But ∠AOC = ∠COB (given)

⇒ 2∠AOC = 180°

⇒ ∠AOC = 90°

According to question

∠1 + ∠2 + ∠3 + ∠4 = 360° (Angles formed at a point is 360°)

But ∠1 + ∠2 + ∠3 = 230° (given)

(∠1 + ∠2 + ∠3) + ∠4 = 360°

230° + ∠4 = 360°

∠4 = 360° – 230°

∠4 = 130°

But ∠1 + ∠4 = 180° (linear pair axiom)

∠1 + 130°= 180°

∠1 = 180°- 130° = 50°

As we know that if a number of lines meet at a point then the sum of all the angles so formed = 360°

Here

∠PAQ + ∠QAB + reflex ∠PAB = 360°

⇒ 55° + ∠QAB + 255° = 360°

⇒ ∠QAB = 360° – 310°

⇒ ∠QAB = 50°

⇒ ∠QAB = ∠ABR = 50° (alt. angle)

QA || BR

Also ∠PAB = ∠ABS = 105° (alt. angles)

PA || BS

Hence QA || BR and PA || BS.

∠PAC + ∠QCA = 75° + 105° = 180° (Sum of the interior angles)

n || p

Also ∠ACD = ∠CDS = 114° (alternate angles)

m || r

Hence n || p and m || r

Here ∠ACB + ∠ACD = 180° (Linear pair of angles)

⇒ 5x + 7y = 180°

⇒ 12y = 180°

⇒ y = 15°

Also x + 3y = 7y (Exterior angle property)

⇒ x = 4y

x = 4 x 15° 

⇒ x = 60°

Correct option is (B) 20°

Let both angles are 2x and 7x.

\(\because\) Both angles are complementary angles to each other.

\(\therefore\) 2x+7x = \(90^\circ\)

\(\Rightarrow\) 9x = \(90^\circ\)

\(\Rightarrow\) x = \(\frac{90^\circ}9=10^\circ\)

\(\therefore\) Required angle = 2x \(=20^\circ\)

Correct option is B) 20°

Correct option is (C) ∠1 + ∠2 > 180°

\(\because\) \(l\) & m intersects in opposite direction of \(\angle1\;\&\;\angle2.\)

\(\therefore\) \(\angle1+\angle2>180^\circ\)

C) ∠1 + ∠2 > 180°

∠POS + ∠SOR + ∠ROQ = 180° (Straight angle)

(x + 45°) + x + (x + 30°) = 180°

⇒ 3x + 75° = 180°

⇒ 3x = 105°

⇒ x = 35°

∠POS = (x + 45°) 

= (35° + 45°) = 80°

∠POM and ∠QOM form a linear pair.

x + y = 180° …(i)

Also x – 2y = 30° (given) …(ii)

Solving (i) and (ii), we get x = 130° and y = 50°

(A) 60°, 84° 

Let the angles be x, x + d and x + 2d 

Given that x = 36° 

∴ the angles are 36°, (36 + d)° and (36 + 2d)° 

Now, 36 + 36 + d + 36 + 2d = 180°

⇒ 3d = 72

⇒ d = 24 

∴ The angels are (36 + 24)° and (36 + 48)°

i.e. 60° and 84°

Correct option is (B) 1

(2, 2) is a solution of \(3x+ay=8\)

\(\Rightarrow\) \(3\times2+a\times2=8\)

\(\Rightarrow\) 2a = 8 - 6 = 2

\(\Rightarrow\) a = \(\frac22\) = 1

Correct option is  B) 1

Correct option is (C)

Adjacent angles are two angles that have a common side and a common vertex but do not overlap in any way.

It is clear that both angles given in option (C) do not have any common side.

Thus, both angles are not adjacent angles.

Correct option is  C)

(i) ∠a, ∠b are not adjacent angles because they have no common vertex.

(ii) ∠a, ∠b are not adjacent angles because they have no common vertex. .

(iii) ∠a, ∠b are adjacent angles

(iv) ∠a, ∠b are adjacent angles.

Correct option is (C) ∠1, ∠7

(\(\angle1\) & \(\angle7\)) and (\(\angle2\) & \(\angle6\)) are pairs of alternative - exterior angles.

Correct option is  C) ∠1, ∠7

Yes, two acute angels can be complement to each other if their sum is 90°.

(i) 50° + 130° = 180° supplement angles.

(ii) 105° + 75° = 380° supplement angles.

(iii) 45° + 100° = 145° * 180° not supplement angles.

(i) In figure, l ॥ m and L in a transversal line

∴ ∠x = 70°(alternate angle)

(ii) In figure ∵ L₁ ≠ L₂

∴ ∠1 ≠ ∠2

Let one angles is x°,

∴ its complementary’ angle = x°

∴ x° + x° = 90°

⇒ 2x° = 90° ⇒ x° = 90°/2 = 45°

Required pair of angles = 45°, 45°

(i) ∠1 and ∠7, ∠4 and ∠6 are pair of alternate angles.

(ii) ∠2 and ∠8, ∠3 and ∠5 are pair of alternate angles.

(iii) ∠1 and ∠5, ∠2 and ∠6, ∠4 and ∠8, ∠3 and ∠7 are pair of corresponding angles.

(iv) ∠1 and ∠6, ∠4 and ∠7 are pair of interior angles on same side of transversal.

Figure (i) and (ii) are pair of parallel lines.

There are following examples from our surroundings where lines intersecting at right angles are:

(i) Edge of the black board.

(ii) Angle between top of the table with its legs.

(iii) Edges of the carrum board.

(iv) Edge of the paper sheet.

Let x° be the supplementary angle of 90°

∴ Sum of supplementary angles = 180°.

∴ 90° + x°= 180°

⇒ x° = 180° – 90° = 90

Required supplement angle = 90°

0° is the complement angle of 90°.

(i) Similar supplementary angles are ∠QOT and ∠TOP.

(ii) Dissimilar supplementary angles are ∠QOS and ∠SOP and ∠POR and ∠ROQ.

(iii) Vertically opposite angles ∠ROQ and ∠SOP.

(iv) Adjacent angles are ∠QOS and ∠SOT, ∠SOT and ∠TOP and ∠TOP and ∠POR.

(v) Adjacent complementary angles arc ∠QOS and ∠SOT.

(i) ∠POQ or ∠QOR
(ii) ∠POQ or ∠QOS
(iii) ∠POR or ∠ROS
(iv) ∠QOR or ∠ROS

An infinite number of transversals can be drawn for a pair of lines.

(i) ∠3 and ∠4 are alternate angles.

(ii) ∠7 and ∠8 are alternate angles.

(iii) ∠1 and ∠2 are linear pair.

(iv) ∠5 and ∠6 are internal angles.

(v) ∠9 and ∠10 are corresponding angles.

(i) 40° + 50° = 90° are complement angles,

(ii) 30° + 50° = 80° 90° not complement angles,

(iii) 75° + 15° = 90° complement angles.

(i) If external angle is a right angle then sum of internal opposite angle will also be 90°. it means each angle will be an acute angle.

(ii) If external angle is an obtuse angle then out of two internal opposite angles either both are acute angle or one is obtuse angle and other is acute angle.

(iii) If external angle is an acute angle than each internal opposite angle will be acute angle.

Let x is the complement of 45°

Then x + 45° = 90°

⇒ x = 90° – 45° = 45°

∴ Complement angle of 45° = 45°

No, two obtuse angles can not be complement to each other because their sum is never 90°.

We know that sum of two complementary angles are 90°. Therefore

(i) Measure of complementary angle of 45°
= (90° – 45°) = 45°

(ii) Measure of complementary angle of 65°
= (90° – 65°) = 25°

(iii) Measure of complementary angle of 41°
= (90° – 41°) = 49°

(iv) Measure of complementary angle of 54°
= (90° – 54°) = 36°

We know that sum of two supplementary angles are 180°. Therefore

(i) Supplementary angle of 100°
= (180° – 100°) = 80°

(ii) Supplementary angle of 90°
= (180° – 90°) = 90°

(iii) Supplementary angle of 55°
= (180° – 55°) = 125°

(iv) Supplementary angle of 125°
= (180° – 125°) = 55°

Sum of complements of each other is 90°

(i) Complement angle of 20° = 90°- 20° = 70°

(ii) Complement angle of 63° = 90° – 63° = 27°

(iii) Complement angle of 57° = 90° – 57° = 33°

(i) one

(ii) different

(iii) concurrent

(iv) equal

(i) 140°, 40°

Sum of angles = 140° + 40° = 180°

∴ These are supplementary angles.

(ii) 170°, 10°

Sum of angles = 170° + 10° = 180°

∴ These are supplementary angles.

(iii) 75°, 15°

Sum of angles = 75° + 15° = 90°

∴ These are complementary angles.

(iv) 33°, 57°

Sum of angles = 33° + 57° = 90°

∴ These are complementary angles.

(v) 115°, 65°

Sum of angles = 115° + 65° – 180°

∴ These are supplemental angles.

(vi) 25°, 65°

Sum of angles = 25° + 65° = 90°

∴ It is complementary angles.

In figure (i) p ॥ q

∴ ∠x = 95°

and ∠x + 30° + ∠y= 180° (Linear pair)

95°+ 30 ° + ∠y = 180°

125° + ∠y = 180°

∠y = 180°- 125° = 55°

In figure (ii) p ॥ q and q ॥ r

∴ ∠y = 65° (Alternate angles)

and ∠x + ∠y = 180° (Linear pair)

∠x + 65° = 180°

∠x = 180° -65°= 115°

The moment of inertia increases & angular velocity decreases.

This is based on the principle of conservation of angular momentum.

An ice-skater, a ballet dancer or an acrobat is able to change his/her angular speed during the course of his/her performance. When the performer stretches out his/her hands and legs, his/her moment of inertia folds his/her hands and legs near his/her body, the moment of inertia decreases and he/she is able to increase his/her angular speed.

Surface tension of water decreases when soap is dissolved in it so spraying becomes easier became less work is required to spray it.

Water, because surface tension of water is much smaller as compared to that of mercury.

(a) caffeine