COMPLEMENTARY AND SUPPLEMENTARY ANGLES PROBLEMS AND SOLUTIONS

Problem 1 :

Find the measure of an angle that is twice as large as its supplement.

Solution :

Let x be the measure of the required angle.

The measure of a supplement of the angle x is (180° - x).

Given : The measure of an angle that is twice as large as its supplement.

x = 2(180° - x)

x = 360° - 2x

Add 2x to both sides.

3x = 360°

Divide both sides by 3.

x = 120°

Therefore, the required angle is 120°.

Problem 2 :

Find the measure of an angle that is half as large as its complement.

Solution :

Let x be the measure of the required angle.

The measure of a complement of the angle x is (90° - x).

Given : The measure of an angle that is half as large as its complement.

Multiply both sides by 2.

2x = 90° - x

Add x to both sides.

3x = 90°

Divide both sides by 3.

x = 30°

Therefore, the required angle is 30°.

Problem 3 :

The measure of a supplement of an angle is 12 more than twice the measure of the angle. Find the measures of the angle and its supplement.

Solution :

Let x be the measure of the required angle.

The measure of a supplement of the angle x is (180° - x).

Given : The measure of a supplement of an angle is 12 more than twice the measure of the angle.

180° - x = 2x + 12°

Add x to both sides.

180° = 3x + 12°

Subtract 12 from both sides.

168° = 3x

Divide both sides by 3.

56° = x

And also,

180° - x = 180° - 56° = 124°

Therefore, the measure of the angle is 56° and its supplement is 124°.

Problem 4 :

A supplement of an angle is six times as large as a complement of the angle. Find the measures of the angle, its supplement and its compliment.

Solution :

Let x be the measure of the required angle.

The measure of a complement and a supplement of the angle x are (90° - x) and (180° - x) respectively.

Given : A supplement of an angle is six times as large as a complement of the angle.

180° - x = 6(90° - x)

180° - x = 540° - 6x

Add 6x to both sides.

180° + 5x = 540°

Subtract 180° from both sides.

5x = 360°

Divide both sides by 5.

x = 72°

Therefore, the measure of the angle is 72°, its supplement is 108° and its complement is 18°.

Problem 5 :

You are told that the measure of an acute angle is equal to the difference between the measure of a supplement of the angle and twice the measure of a complement of the angle. What can you deduce about the angle? Explain.

Solution :

Let x be the measure of an acute angle.

The measure of a complement and a supplement of the angle x are (90° - x) and (180° - x) respectively.

Given : The measure of an acute angle is equal to the difference between the measure of a supplement of the angle and twice the measure of a complement of the angle.

x = (180° - x) - 2(90° - x)

x = 180° - x - 180° + 2x

x = x

The above is equation is true for all values of x such that x is an acute angle.

Conclusion :

The measure of any acute angle is equal to the difference between the measure of a supplement and twice the measure of a complement of the acute angle.

Problem 6 :

Can the measure of a complement of an angle ever equal exactly half the measure of a supplement of the angle? Explain. 

Solution :

Let x be the measure of the angle.

The measure of a complement and a supplement of the angle x are (90° - x) and (180° - x) respectively.

Given : The measure of a complement of an angle ever equal exactly half the measure of a supplement of the angle.

Multiply both sides by 2.

2(90° - x) = 180° - x

180° - 2x = 180° - x

Subtract 180° from both sides.

-2x = -x

Add x to both sides.

-x = 0°

Multiply both sides by -1.

x = 0°

Conclusion :

Yes, the measure of a complement of an angle ever equal exactly half the measure of a supplement of the angle and that angle is 0°. 

Problem 7 :

The m<A is complementary to the <B. The m<C is complementary to m<B. If m<A = 62°, what is m<B and the m<C.

Solution :

Since <A and <B are complementary angles,

<A + <B = 90°

Given that, m<A = 62° 

62° + <B = 90

<B = 90 - 62

<B = 18°

Since m<B and m<C are complementary, then 

18° + m<C = 90

m<C = 90 - 18

m<C = 62°

Problem 8 :

The m<D is supplementary to the m<E. The m<F is supplementary to m<E. If m<F = 113°, what is m<D and the m<E.

Solution :

Since <D and <E are supplementary angles,

m<D + m<E = 180° -----(1)

m<F + m<E = 180°-----(2)

Applying m<F = 113° in (2), we get

113° + m<E = 180°

m<E = 180° - 113°

m<E = 67°

By applying m<E = 67° in (1), we get

m<D + 67° = 180°

m<D = 180° - 67°

m<D = 113°

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