COMPLEMENTARY AND SUPPLEMENTARY ANGLES WORD PROBLEMS

In this section, you will learn how to solve word problems on complementary and supplementary angles.

Problem 1 :

Two angles are complementary. If one of the angles is double the other angle, find the two angles. 

Solution :

Let x be one of the angles. 

Then the other angle is 2x.

Because x and 2x are complementary angles, we have

x + 2x  =  90°

3x  =  90

Divide each side by 3. 

x  =  30

And,

2x  =  2(30)  =  60

So, the two angles are 30° and 60°.

Problem 2 :

Two angles are complementary. If one angle is two times the sum of other angle and 3, find the two angles.

Solution :

Let x and y be the two angles which are complementary.

So, we have

x + y  =  90° -----> (1)

Given : One angle is two times the sum of other angle and 3. 

Then, 

x  =  2(y + 3)

x  =  2y + 6 ----->(2)

Now, substitute (2y + 6) for x in (1). 

(1)-----> 2y + 6 + y  =  90

3y + 6  =  90

Subtract 6 from each side. 

3y  =  84

Divide each side by 3.

y  =  28

Substitute 28 for y in (2).

(2)-----> x  =  2(28) + 6

x  =  56 + 6

x  =  62

So, the two angles are 62° and 28°.

Problem 3 :

The measure of an angle is 3/4 of 60°. What is the measure of the complementary angle ? 

Solution :

Let x be the measure of a complementary angle required.

Given : The measure of an angle is 3/4 of 60°. 

Then, 

3/4 ⋅ 60°  =  45° 

Because x and 45° are complementary angles, we have

x + 45  =  90

Subtract 45° from each side. 

x  =  45

So, the measure of the complementary angle is 45°.

Problem 4 :

Two angles are supplementary. If one angle is 36° less than twice of the other angle, find the two angles. 

Solution :

Let x and y be the two angles which are supplementary.

Then,

x + y  =  180° ----->(1)

From the information, "one angle is 36

Given : One angle is 36° less than twice of the other angle.

Then,

x  =  2y - 36 ----->(2)

Now substitute (2y - 6) for x in (1), 

(1)-----> 2y - 36 + y  =  180

3y - 36  =  180

Add 36 to each side. 

3y  =  216

Divide each side by 3. 

y  =  72

Now, Substitute 72 for y in (2).

(2)-----> x  =  2(72) - 36

x  =  144 - 36

x  =  108

So, the two angles are 108° and 72°.

Problem 5 :

An angle and its one-half are complementary. Find the angle.

Solution :

Let x be the required angle.

Then, one-half of the angle is x/2. 

Given : An angle and its one-half are complementary. 

Then,

x + x/2  =  90°

2x/2 + x/2  =  90

(2x + x) / 2  =  90

3x/2  =  90

Multiply each side by 2.

3x  =  180

Divide each side by 3.

x  =  60

So, the required angle is 60°.

Problem 6 :

Two angles are supplementary. If 5 times of one angle is 10 times of the other angle. Find the two angles.

Solution :

Let x and y be the two angles which are supplementary.

Then, 

x + y  =  180 -----(1)

Given : 5 times of one angle is 10 times of the other angle. 

Then,

5x  =  10y

Divide each side by 5. 

x  =  2y ----(2)

Substitute 2y for x in (1).

2y + y  =  180

3y  =  180

Divide each side by 3.

y  =  60

Substitute 60 for y in (2).

x  =  2(60)

x  =  120

So, the two angles are 60° and 120°.

Problem 7 :

The measure of the supplement of angle A is 40 degrees larger than twice the measure of the complement of angle of A. What is the sum in degrees, of the measures of the supplement and complement of angle A ? 

Solution :

Let x be the measure of angle A.

Then,

supplement of angle A  =  180° - x 

complement of angle A  =  90° - x 

Given : The measure of the supplement of angle A is 40 degrees larger than twice the measure of the complement of angle of A. 

That is, 

(180° - x) is 40° more than 2(90° - x)

180 - x  =  2(90 - x) + 40

180 - x  =  180 - 2x + 40

Subtract 180 from each side. 

- x  =  - 2x + 40

Add 2x to each side.

x  =  40

Then, supplement of angle A is

=  180 - x

=  180 - 40

=  140°

Complement of angle A is

=  90 - x

=  90 - 40

=  50°

The sum of supplement and complement of angle A is 

=  140° + 50° 

=  190°

So, the sum of supplement and complement of angle A is 190° degrees. 

Problem 8 :

Twice the complement of an angle is 24 degrees less than its supplement. What is the measure of the angle ?

Solution :

Let x be the measure of angle.

Then,

supplement of angle  =  180° - x 

complement of angle  =  90° - x 

Given : Twice the complement of an angle is 24 degrees less than its supplement.

That is, 

2(90° - x) is 24 degrees less than (180° - x)   

Then,

2(90 - x)  =  (180 - x) - 24

180 - 2x  =  180 - x - 24

Subtract 180 from each side. 

- 2x  =  - x - 24

Add x to each side. 

- x  =  - 24

Multiply each side by (-1).

x  =  24

So, the measure of the angle is 24 degrees. 

Problem 9 :

Two angles are supplementary. If the ratio of the measure of the smaller angle to that of the larger angle is 5:7. What is the measure of the smaller angle ? 

Solution :

Given : The ratio of the measure of the smaller angle to that of the larger angle is 5 : 7.

Then,

measure of the smaller angle  =  5x

measure of the larger angle  =  7x

Given : Two angles are supplementary.

Then,

5x  +  7x  =  180

12x  =  180

Divide each side by 12.

x  =  15

Measure of the smaller angle is 

=  5x

=  5(15)

=  75°

So, the measure of the smaller angle is 75°

Problem 10 :

The measure of supplement of an angle is equal to the twice the measure of the angle. What is the measure in degrees, of the compliment of the angle ?

Solution :

Let x be the measure of the angle.

Then, 

measure of supplement of the angle  =  180° - x

Given : The measure of supplement of the angle is equal to the twice the measure of the angle.

That is,

(180° - x) is equal to 2x

Then, 

180 - x  =  2x

Add x to each side. 

180  =  3x

Divide each side by 3. 

60  =  x

So, the measure of the angle is 60°.

Then, the measure of complement of the angle is

=  90 - 60

=  30°

So, the measure of compliment of  the angle is 30 degrees. 

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