Class 9th 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 ΔABC, ∠A + ∠B = 70° and ∠B + ∠C = 135°. Find the measure of each angle of the…
- In the given figure at right, side BC of ΔABC 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 ∠…
- 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 ΔABC, ∠A + ∠B = 70° and ∠B + ∠C = 135°. Find the measure of each angle of the…
- In the given figure at right, side BC of ΔABC is produced to D. Find ∠A and ∠C.…
- 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 l1 || l2 and m1 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 l1 and l2 be parallel lines.
i. 
ii. 
Answer:(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°
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 ΔABC, ∠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 ΔABC 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°
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 l1 || l2 and m1 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 l1 and l2 be parallel lines.
i.
ii.
Answer:
(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°
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 ΔABC, ∠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 ΔABC 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°.
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°.