Before we solve the workedout problems on complementary and supplementary angles we will recall the definition of complementary angles and supplementary angles.
Complementary Angles:
Two angles are called complementary angles, if their sum is one right angle i.e. 90°.
Each angle is called the complement of the other.
Example, 20° and 70° are complementary angles, because 20° + 70° = 90°.
Clearly, 20° is the complement of 70° and 70° is the complement of 20°.
Thus, the complement of angle 53° = 90°  53° = 37°.
Supplementary Angles:
Two angles are called supplementary angles, if their sum is two right angles i.e. 180°.
Each angle is called the supplement of the other.
Example, 30° and 150° are supplementary angles, because 30° + 150° = 180°.
Clearly, 30° is the supplement of 150° and 150° is the supplement of 30°.
Thus, the supplement of angle 105° = 180°  105° = 75°.
Solved problems on complementary and supplementary angles:
1. Find the complement of the angle 2/3 of 90°.
Solution:
Convert 2/3 of 90°
2/3 × 90° = 60°
Complement of 60° = 90°  60° = 30°
Therefore, complement of the angle 2/3 of 90° = 30°
2. Find the supplement of the angle 4/5 of 90°.
Solution:
Convert 4/5 of 90°
4/5 × 90° = 72°
Supplement of 72° = 180°  72° = 108°
Therefore, supplement of the angle 4/5 of 90° = 108°
3. The measure of two complementary angles are (2x  7)° and (x + 4)°. Find the value of x.
Solution:
According to the problem, (2x  7)° and (x + 4)°, are complementary angles’ so we get;
(2x  7)° + (x + 4)° = 90°
or, 2x  7° + x + 4° = 90°
or, 2x + x  7° + 4° = 90°
or, 3x  3° = 90°
or, 3x  3° + 3° = 90° + 3°
or, 3x = 93°
or, x = 93°/3°
or, x = 31°
Therefore, the value of x = 31°.
4. The measure of two supplementary angles are (3x + 15)° and (2x + 5)°. Find the value of x.
Solution:
According to the problem, (3x + 15)° and (2x + 5)°, are complementary angles’ so we get;
(3x + 15)° + (2x + 5)° = 180°
or, 3x + 15° + 2x + 5° = 180°
or, 3x + 2x + 15° + 5° = 180°
or, 5x + 20° = 180°
or, 5x + 20°  20° = 180°  20°
or, 5x = 160°
or, x = 160°/5°
or, x = 32°
Therefore, the value of x = 32°.
5. The difference between the two complementary angles is 180°. Find the measure of the angle.
Solution:
Let one angle be of measure x°.
Then complement of x° = (90  x)
Difference = 18°
Therefore, (90°  x) – x = 18°
or, 90°  2x = 18°
or, 90°  90°  2x = 18°  90°
or, 2x = 72°
or, x = 72°/2°
or, x = 36°
Also, 90°  x
= 90°  36°
= 54°.
Therefore, the two angles are 36°, 54°.
6. POQ is a straight line and OS stands on PQ. Find the value of x and the measure of ∠ POS, ∠ SOR and ∠ ROQ.
Solution:
POQ is a straight line.
Therefore, ∠POS + ∠SOR + ∠ROQ = 180°
or, (5x + 4°) + (x  2°) + (3x + 7°) = 180°
or, 5x + 4° + x  2° + 3x + 7° = 180°
or, 5x + x + 3x + 4°  2° + 7° = 180°
or, 9x + 9° = 180°
or, 9x + 9°  9° = 180°  9°
or, 9x = 171°
or, x = 171/9
or, x = 19°
Put the value of x = 19°
Therefore, x  2
= 19  2
= 17°
Again, 3x + 7
= 3 × 19° + 7°
= 570 + 7°
= 64°
And again, 5x + 4
= 5 × 19° + 4°
= 95° + 4°
= 99°
Therefore, the measure of the three angles is 17°, 64°, 99°.
These are the above solved examples on complementary and supplementary angles explained stepbystep with detailed explanation.
● Lines and Angles
Fundamental Geometrical Concepts
Some Geometric Terms and Results
Complementary and Supplementary Angles
Parallel and Transversal Lines
7th Grade Math Problems
8th Grade Math Practice
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