Geometry

i. We know that two angles are said to be complementary if their sum is 90°.

So, the complement of 63° = 90° - 63° = 27°

ii. We know that two angles are said to be complementary if their sum is 90°.

So, the complement of 24° = 90° - 24° = 66°

iii. We know that two angles are said to be complementary if their sum is 90°.

So, the complement of 48° = 90° - 48° = 42°

iv. We know that two angles are said to be complementary if their sum is 90°.

So, the complement of 35° = 90° - 35° = 55°

v. We know that two angles are said to be complementary if their sum is 90°.

So, the complement of 20° = 90° - 20° = 70°

i. We know that two angles are said to be supplementary if their sum is 180°.

So, the supplement of 58° = 180° - 58° = 122°

ii. We know that two angles are said to be supplementary if their sum is 180°.

So, the supplement of 148° = 180° - 148° = 32°

iii. We know that two angles are said to be supplementary if their sum is 180°.

So, the supplement of 120° = 180° - 120° = 60°

iv. We know that two angles are said to be supplementary if their sum is 180°.

So, the supplement of 40° = 180° - 40° = 140°

v. We know that two angles are said to be supplementary if their sum is 180°.

So, the supplement of 100° = 180° - 100° = 80°

i. In the given figure,

AB is a straight line, hence all the angles formed on it are supplementary.

⇒ (x – 20)° + x° + 40° = 180°

⇒ 2x + 20° = 180°

⇒ 2x = 180° - 20°

⇒ 2x = 160°

⇒ x = 80°

ii. In the given figure,

AB is a straight line, hence all the angles formed on it are supplementary.

⇒ (x + 30)° + (115 – x)° + x° = 180°

⇒ x + 145° = 180°

⇒ x = 180° - 145°

⇒ x = 35°

Let the complement of an angle be x°

According to the question,

Angle = 2x°

We know that two angles are said to be complementary if their sum is 90°.

So, 2x° + x° = 90°

⇒ 3x° = 90°

⇒ x° = 30°

∴ Angle = 2x° = 2× 30° = 60°, Compliment = x° = 30°

Let the supplement of an angle be x°

According to the question,

Angle = 4x°

We know that two angles are said to be supplementary if their sum is 180°.

So, 4x° + x° = 180°

⇒ 5x° = 180°

⇒ x° = 36°

∴ Angle = 4x° = 4× 36° = 144°, Compliment = x° = 36°

Let the compliment of an angle be x°

According to the question,

Supplement of the angle = 4x°

We know that two angles are said to be supple mentary if their sum is 180°.

So, required angle = 180° - 4x° ..(I)

Also, if the two angles are complimentary their sum is 90°.

⇒ required angle = 90° - x° …(II)

Equating I and II,

90° - x° = 180° - 4x°

⇒ 3x° = 90°

⇒ x° = 30°

Hence, required angle = 90° - x° = 90° - 30° = 60°

Let the supplement of an angle be x°

According to the question,

Compliment of the angle

We know that two angles are said to be supple-mentary if their sum is 180°.

So, ..(I)

Also, if the two angles are complimentary their sum is 90°.

…(II)

Equating I and II,

⇒ x° = 108°

Hence, required angle = 180° - x° = 180° - 108° = 72°

Let the supplementary angles be 4x and 5x

We know that two angles are said to be supple mentary if their sum is 180°.

⇒ 4x + 5x = 180°

⇒ 9x = 180°

⇒ x = 20°

So, the supplementary angles will be 4× 20° = 80° and 5× 20° = 100°

Let the complimentary angles be 3x and 2x

We know that two angles are said to be complimentary if their sum is 90°.

⇒ 2x + 3x = 90°

⇒ 5x = 90°

⇒ x = 18°

So, the complimentary angles will be 2×18° = 36° and 3×18° = 54°

Given: ∠ADC = 3x and ∠BDC = 2x

Here AB is the straight line.

∴ ∠ADC and ∠BDC are linear pair.

⇒ ∠ADC + ∠BDC = 180°

⇒ 3x + 2x = 180°

⇒ 5x = 180°

⇒ x = 36°

Given: ∠ADC = (3x + 5)° and ∠BDC = (2x – 25)°

Here AB is the straight line.

∴ ∠ADC and ∠BDC are linear pair.

⇒ ∠ADC + ∠BDC = 180°

⇒ 3x + 5 + 2x - 25 = 180°

⇒ 5x - 20 = 180°

⇒ 5x = 200°

⇒ x = 40°

Given: ∠AOD = y° ,∠DOC = 90°,∠COB = x° ,∠AOE = 3x and ∠BOE = 60°

Here AB is the straight line.

∴ ∠AOE and ∠BDC are linear pair.

⇒ ∠ADC + ∠BOE = 180°

⇒ 3x + 60° = 180°

⇒ 3x = 120°

⇒ x = 40°

Also, ∠AOD, ∠DOC and ∠BOC are linear pair.

⇒ ∠AOD + ∠DOC + ∠BOC = 180°

⇒ y + 90° + x° = 180°

⇒ y = 90° - 40°

⇒ y = 50°

Given: ∠F = 65°

∠B = ∠F = 65° {Corresponding angle}

∠H = ∠F = 65° {vertically opposite angle}

∠D = ∠B = 65° {vertically opposite angle}

Also, ∠C + ∠F = 180° {co-interior angles are supplementary}

⇒ ∠C = 180° - 65°

⇒ ∠C = 115°

∠E = ∠C = 115° {Alternate interior angle}

∠G = ∠C = 115° {Corresponding angle}

∠A = ∠G = 115° {Alternate exterior angle}

(i) Given: l₁||l₂

∠1 = (2x + 20)° and ∠2 = (3x-10)°

∵ Corresponding angles are equal

∴ ∠1 = ∠2

⇒ (2x + 20)° = (3x-10)°

⇒ x = 30°

ii. Given: l₁||l₂

∠1 = (2x)° and ∠2 = (3x + 20)°

∵ Co-interior angles are supplementary

∴ ∠1 + ∠2 = 180°

⇒ (2x)° + (3x + 20)° = 180°

⇒ 5x = 160°

⇒ x = 32°

Let the angles of a triangle be x, 2x and 3x.

We know that sum of the angles of a triangle is 180°.

⇒ x + 2x + 3x = 180°

⇒ 6x = 180°

⇒ x = 30°

So, the angles of the triangle are 30°, 2×30° = 60° and 3× 30° = 90°.

Given ∠A + ∠B = 70° and ∠B + ∠C = 135°

We know that sum of the angles of a triangle is 180°.

∠A + ∠B + ∠C = 180° …I

Also, ∠A + ∠B + ∠B + ∠C = 70° + 135°

⇒ (∠A + ∠B + ∠C) + ∠B = 205°

From I,

⇒ 180° + ∠B = 205°

⇒ ∠B = 205° - 180°

⇒ ∠B = 25°

Putting in given equations,

∠A + ∠B = 70° and ∠B + ∠C = 135°

⇒ ∠A + 25° = 70° and 25° + ∠C = 135°

⇒ ∠A = 45° and ∠C = 110°

So, the angles of the triangle are 45°, 25° and 110°.

Given ∠B = 40° and ∠ACD = 120°

∵ ∠ACD is an external angle

∴ ∠ACD = ∠A + ∠B

⇒ ∠A + 40° = 120°

⇒ ∠A = 80°

We know that sum of the angles of a triangle is 180°.

∠A + ∠B + ∠C = 180°

⇒ 80° + + 40° + ∠C = 180°

⇒ ∠C = 180° - 120°

⇒ ∠C = 60°

Let the required angle be x and the supplement of x is y.

As we know if two angles are supplement of each other, then sum of those two angles is 180°.

⇒ x + y = 180

⇒ y = 180 – x

Now, according to question;

Required angle is equal to one third of its supplement;

i.e.

⇒ y = 3x

Now putting the value of y in this equation,

⇒ 180 – x = 3x

⇒

⇒

⇒

Hence, required angle is 45°.

Hence, correct option is (c).

Suppose that

BOC = 2x and ∠ AOC = 2y

Now, by question; OP and OQ is the bisect or of ∠ BOC and ∠ AOC respectively.

⇒

Similarly,

Now,

∠ POQ = ∠ POC + ∠ COQ = x + y

So, we have to find ∠ POQ i.e.x + y Now, ∠ BOC + ∠ AOC = 180

⇒ 2x + 2y = 180

⇒ 2(x + y) = 180

⇒ x + y = 180/2

⇒ x + y = 90°

Hence, ∠ POQ = 90°.

Let the required angle be x and it’s complement is y.

So,

⇒ y = 90 – x

Now, according to question;

Complement of the angle exceeds the angle by 60° i.e. y exceeds the x by 60°.

⇒ y – 60 = x

Now, putting the value of your in this equation,

⇒ (90 – x) – 60 = x

⇒ 30 – x = x

⇒ 2x = 30

⇒ x = 15°

Hence, required angle is 15°.

Let the required angle be x and it’s complement is y and it’s supplement is z.

⇒ x + y = 90

i.e. y = 90 – x

And, x + z = 180

i.e. z = 180 – x

Now, according to question,

Six times of complement of x is 12°

less than twice of its supplement.

i.e. 6y = 2z – 12

Putting the value of y and z in this equation,

⇒ 6 (90 – x) = 2 (180 – x) – 12

⇒ 540 – 6x = 360 – 2x – 12

⇒ 6x – 2x = 540 – 360 + 12

⇒ 4x = 192

⇒

⇒ x = 48°

Hence required angle is 48°.

Let ∠ B and ∠ C be 2x and 3x respectively.

Now, As we know sum of two interior angles of a triangle is equal to third exterior angle;

⇒ ∠ B + ∠ C = 130°

⇒ 2x + 3x = 130°

⇒ 5x = 130°

⇒ x = 26°

So, ∠ B = 2x = 2×26° = 52°

Hence, required angle is 52°.

Given: ABCD is a parallelogram and E is the mid – point of AB such that CE bisects ∠ BCD.

construction: Join DE. The figure is given below:

Let ∠ BCD = 2x

⇒ ∠ BCE = ∠ ECD = ∠ BCD/2

Now, let ∠ ADC = 2y

And as we joined DE it will also bisect ∠ ADC.

Therefore, ∠ EDC =

Now, as we know sum of adjacent angles of parallelogram is equal to 180°.

⇒ ∠ BCD + ∠ ADC = 180°

⇒ 2x + 2y = 180°

⇒ 2 (x + y) = 180°

⇒ x + y = 90°

Now, In triangle CED,

∠ CED + ∠ EDC + ∠ DCE = 180°

⇒ ∠ CED + x + y = 180°

⇒ ∠ CED + 90° = 180°

⇒ ∠ CED = 180° – 90°

⇒ ∠ CED = 90°

Hence required angle is 90°.