##### Class 9^{th} Mathematics Term 1 Tamilnadu Board Solution

**Exercise 4.1**- Find the complement of each of the following angles. i. 63° ii. 24° iii. 48° iv.…
- Find the supplement of each of the following angles. i. 58° ii. 148° iii. 120°…
- Find the value of x in the following figures. i. ii.
- The angle which is two times its complement. Find the angles in each of the…
- The angle which is four times its supplement. Find the angles in each of the…
- The angles whose supplement is four times its complement. Find the angles in…
- The angle whose complement is one sixth of its supplement. Find the angles in…
- Supplementary angles are in the ratio 4:5. Find the angles in each of the…
- Two complementary angles are in the ratio 3:2. Find the angles in each of the…
- Find the values of x, y in the following figures.
- Find the values of x, y in the following figures.
- Find the values of x, y in the following figures.
- Let l1 || l2 and m1 is a transversal. If ∠F = 65°, find the measure of each of…
- For what value of x will l1 and l2 be parallel lines. i. ii.
- The angles of a triangle are in the ratio of 1:2:3. Find the measure of each…
- In Î”ABC, ∠A + ∠B = 70° and ∠B + ∠C = 135°. Find the measure of each angle of the…
- In the given figure at right, side BC of Î”ABC is produced to D. Find ∠A and ∠C.…

**Exercise 4.3**- If an angle is equal to one third of its supplement, its measure is equal toA.…
- In the given figure, OP bisect ∠ BOC and OQ bisect ∠ AOC. Then ∠ POQ is equal to…
- The complement of an angle exceeds the angle by 60°. Then the angle is equal…
- Find the measure of an angle, if six times of its complement is 12° less than…
- In the given figure, ∠ B:∠ C = 2:3, find ∠ B A. 120° B. 52° C. 78° D. 130°…
- ABCD is a parallelogram, E is the mid - point of AB and CE bisects ∠ BCD. Then ∠…

**Exercise 4.1**

- Find the complement of each of the following angles. i. 63° ii. 24° iii. 48° iv.…
- Find the supplement of each of the following angles. i. 58° ii. 148° iii. 120°…
- Find the value of x in the following figures. i. ii.
- The angle which is two times its complement. Find the angles in each of the…
- The angle which is four times its supplement. Find the angles in each of the…
- The angles whose supplement is four times its complement. Find the angles in…
- The angle whose complement is one sixth of its supplement. Find the angles in…
- Supplementary angles are in the ratio 4:5. Find the angles in each of the…
- Two complementary angles are in the ratio 3:2. Find the angles in each of the…
- Find the values of x, y in the following figures.
- Find the values of x, y in the following figures.
- Find the values of x, y in the following figures.
- Let l1 || l2 and m1 is a transversal. If ∠F = 65°, find the measure of each of…
- For what value of x will l1 and l2 be parallel lines. i. ii.
- The angles of a triangle are in the ratio of 1:2:3. Find the measure of each…
- In Î”ABC, ∠A + ∠B = 70° and ∠B + ∠C = 135°. Find the measure of each angle of the…
- In the given figure at right, side BC of Î”ABC is produced to D. Find ∠A and ∠C.…

**Exercise 4.3**

- If an angle is equal to one third of its supplement, its measure is equal toA.…
- In the given figure, OP bisect ∠ BOC and OQ bisect ∠ AOC. Then ∠ POQ is equal to…
- The complement of an angle exceeds the angle by 60°. Then the angle is equal…
- Find the measure of an angle, if six times of its complement is 12° less than…
- In the given figure, ∠ B:∠ C = 2:3, find ∠ B A. 120° B. 52° C. 78° D. 130°…
- ABCD is a parallelogram, E is the mid - point of AB and CE bisects ∠ BCD. Then ∠…

###### Exercise 4.1

**Question 1.**Find the complement of each of the following angles.

i. 63° ii. 24°

iii. 48° iv. 35°

v. 20°

**Answer:**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°

**Question 2.**Find the supplement of each of the following angles.

i. 58° ii. 148°

iii. 120° iv. 40°

v. 100°

**Answer:**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°

**Question 3.**Find the value of x in the following figures.

i.

ii.

**Answer:**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°

**Question 4.**Find the angles in each of the following.

The angle which is two times its complement.

**Answer:**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°

**Question 5.**Find the angles in each of the following.

The angle which is four times its supplement.

**Answer:**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°

**Question 6.**Find the angles in each of the following.

The angles whose supplement is four times its complement.

**Answer:**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°

**Question 7.**Find the angles in each of the following.

The angle whose complement is one sixth of its supplement.

**Answer:**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°

**Question 8.**Find the angles in each of the following.

Supplementary angles are in the ratio 4:5.

**Answer:**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°

**Question 9.**Find the angles in each of the following.

Two complementary angles are in the ratio 3:2.

**Answer:**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°

**Question 10.**Find the values of x, y in the following figures.

**Answer:**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°

**Question 11.**Find the values of x, y in the following figures.

**Answer:**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°

**Question 12.**Find the values of x, y in the following figures.

**Answer:**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°

**Question 13.**Let l_{1} || l_{2} and m_{1} is a transversal. If ∠F = 65°, find the measure of each of the remaining angles.

**Answer:**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}

**Question 14.**For what value of x will l_{1} and l_{2} be parallel lines.

i.

ii.

**Answer:**(i) Given: l_{1}||l_{2}

∠1 = (2x + 20)° and ∠2 = (3x-10)°

∵ Corresponding angles are equal

∴ ∠1 = ∠2

⇒ (2x + 20)° = (3x-10)°

⇒ x = 30°

ii. Given: l_{1}||l_{2}

∠1 = (2x)° and ∠2 = (3x + 20)°

∵ Co-interior angles are supplementary

∴ ∠1 + ∠2 = 180°

⇒ (2x)° + (3x + 20)° = 180°

⇒ 5x = 160°

⇒ x = 32°

**Question 15.**The angles of a triangle are in the ratio of 1:2:3. Find the measure of each angle of the triangle.

**Answer:**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°.

**Question 16.**In Î”ABC, ∠A + ∠B = 70° and ∠B + ∠C = 135°. Find the measure of each angle of the triangle.

**Answer:**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°.

**Question 17.**In the given figure at right, side BC of Î”ABC is produced to D. Find ∠A and ∠C.

**Answer:**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°

**Question 1.**

Find the complement of each of the following angles.

i. 63° ii. 24°

iii. 48° iv. 35°

v. 20°

**Answer:**

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°

**Question 2.**

Find the supplement of each of the following angles.

i. 58° ii. 148°

iii. 120° iv. 40°

v. 100°

**Answer:**

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°

**Question 3.**

Find the value of x in the following figures.

i.

ii.

**Answer:**

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°

**Question 4.**

Find the angles in each of the following.

The angle which is two times its complement.

**Answer:**

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°

**Question 5.**

Find the angles in each of the following.

The angle which is four times its supplement.

**Answer:**

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°

**Question 6.**

Find the angles in each of the following.

The angles whose supplement is four times its complement.

**Answer:**

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°

**Question 7.**

Find the angles in each of the following.

The angle whose complement is one sixth of its supplement.

**Answer:**

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°

**Question 8.**

Find the angles in each of the following.

Supplementary angles are in the ratio 4:5.

**Answer:**

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°

**Question 9.**

Find the angles in each of the following.

Two complementary angles are in the ratio 3:2.

**Answer:**

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°

**Question 10.**

Find the values of x, y in the following figures.

**Answer:**

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°

**Question 11.**

Find the values of x, y in the following figures.

**Answer:**

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°

**Question 12.**

Find the values of x, y in the following figures.

**Answer:**

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°

**Question 13.**

Let l_{1} || l_{2} and m_{1} is a transversal. If ∠F = 65°, find the measure of each of the remaining angles.

**Answer:**

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}

**Question 14.**

For what value of x will l_{1} and l_{2} be parallel lines.

i.

ii.

**Answer:**

(i) Given: l_{1}||l_{2}

∠1 = (2x + 20)° and ∠2 = (3x-10)°

∵ Corresponding angles are equal

∴ ∠1 = ∠2

⇒ (2x + 20)° = (3x-10)°

⇒ x = 30°

ii. Given: l_{1}||l_{2}

∠1 = (2x)° and ∠2 = (3x + 20)°

∵ Co-interior angles are supplementary

∴ ∠1 + ∠2 = 180°

⇒ (2x)° + (3x + 20)° = 180°

⇒ 5x = 160°

⇒ x = 32°

**Question 15.**

The angles of a triangle are in the ratio of 1:2:3. Find the measure of each angle of the triangle.

**Answer:**

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°.

**Question 16.**

In Î”ABC, ∠A + ∠B = 70° and ∠B + ∠C = 135°. Find the measure of each angle of the triangle.

**Answer:**

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°.

**Question 17.**

In the given figure at right, side BC of Î”ABC is produced to D. Find ∠A and ∠C.

**Answer:**

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°

###### Exercise 4.3

**Question 1.**If an angle is equal to one third of its supplement, its measure is equal to

A. 40°

B. 50°

C. 45°

D. 55°

**Answer:**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).

**Question 2.**In the given figure, OP bisect ∠ BOC and OQ bisect ∠ AOC. Then ∠ POQ is equal to

A. 90°

B. 120°

C. 60°

D. 100°

**Answer:**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°.

**Question 3.**The complement of an angle exceeds the angle by 60°. Then the angle is equal to

A. 25°

B. 30°

C. 15°

D. 35°

**Answer:**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°.

**Question 4.**Find the measure of an angle, if six times of its complement is 12° less than twice of its supplement.

A. 48°

B. 96°

C. 24°

D. 58°

**Answer:**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°.

**Question 5.**In the given figure, ∠ B:∠ C = 2:3, find ∠ B

A. 120°

B. 52°

C. 78°

D. 130°

**Answer:**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°.

**Question 6.**ABCD is a parallelogram, E is the mid – point of AB and CE bisects ∠ BCD. Then ∠ DEC is

A. 60°

B. 90°

C. 100°

D. 120°

**Answer:**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°.

**Question 1.**

If an angle is equal to one third of its supplement, its measure is equal to

A. 40°

B. 50°

C. 45°

D. 55°

**Answer:**

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).

**Question 2.**

In the given figure, OP bisect ∠ BOC and OQ bisect ∠ AOC. Then ∠ POQ is equal to

A. 90°

B. 120°

C. 60°

D. 100°

**Answer:**

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°.

**Question 3.**

The complement of an angle exceeds the angle by 60°. Then the angle is equal to

A. 25°

B. 30°

C. 15°

D. 35°

**Answer:**

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°.

**Question 4.**

Find the measure of an angle, if six times of its complement is 12° less than twice of its supplement.

A. 48°

B. 96°

C. 24°

D. 58°

**Answer:**

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°.

**Question 5.**

In the given figure, ∠ B:∠ C = 2:3, find ∠ B

A. 120°

B. 52°

C. 78°

D. 130°

**Answer:**

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°.

**Question 6.**

ABCD is a parallelogram, E is the mid – point of AB and CE bisects ∠ BCD. Then ∠ DEC is

A. 60°

B. 90°

C. 100°

D. 120°

**Answer:**

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°.