FINDING UNKNOWN ANGLES IN A SUPPLEMENTARY PAIR

Find the value of x in each supplementary angle pair.

Problem 1 :

Solution :

The sum of the measures of supplementary angles = 180˚.

Here, ∠AOC = x and ∠COB = 100˚

∠AOC + ∠COB = 180˚

x + 100˚ = 180˚

x = 180˚ – 100˚

x = 80˚

So, the value of x is 80˚.

Problem 2 :

Solution :

The sum of the measures of supplementary angles = 180˚.

Here, ∠AOC = 76˚ and ∠COB = x

∠AOC + ∠COB = 180˚

 76˚ + x = 180˚

x = 180˚ - 76˚

x = 104˚

So, the value of x is 104˚.

Problem 3 :

Solution :

The sum of the measures of supplementary angles = 180˚.

Here, ∠AOC = 34˚ and ∠COB = x

∠AOC + ∠COB = 180˚

 34˚ + x = 180˚

x = 180˚ - 34˚

x = 146˚

So, the value of x is 146˚.

Problem 4 :

Solution :

The sum of the measures of supplementary angles = 180˚.

Here, ∠AOC = x and ∠COB = 125˚

∠AOC + ∠COB = 180˚

x + 125˚ = 180˚

x = 180˚ - 125˚

x = 55˚

So, the value of x is 55˚.

Problem 5 :

Solution :

The sum of the measures of supplementary angles = 180˚.

Here, ∠AOC = x and ∠COB = 163˚

∠AOC + ∠COB = 180˚

x + 163˚ = 180˚

x = 180˚ - 163˚

x = 17˚

So, the value of x is 17˚.

Problem 6 :

Solution :

The sum of the measures of supplementary angles = 180˚.

Here, ∠AOC = x and ∠COB = 180˚

∠AOC + ∠COB = 180˚

x + 80˚ = 180˚

x = 180˚ - 80˚

x = 100˚

So, the value of x is 100˚.

Problem 7 :

Solution :

The sum of the measures of supplementary angles =180˚.

Here, ∠AOC = 48˚ and ∠COB = x

∠AOC + ∠COB = 180˚

 48˚ + x = 180˚

x = 180˚ - 48˚

x = 132˚

So, the value of x is 132˚.

Problem 8 :

Solution :

The sum of the measures of supplementary angles = 180˚.

Here, <AOC = 107˚ and <COB = x

<AOC + <COB = 180˚

 107˚ + x = 180˚

x = 180˚ - 107˚

x = 73˚

So, the value of x is 73˚.

Problem 9 :

Two angles form a linear pair. The measure of one angle is five times the measure of the other angle. Find the measure of each angle.

Solution :

Let x be the other angle. Then the first angle will be 5x.

It forms a linear pair, then sum of these angles will be 180 degree.

x + 5x = 180

6x = 180

x = 180/6

x = 30

5x = 5(30) ==> 150

So, the required angles are 30 and 150.

Problem 10 :

Find the measure of each angle, ∠EFG and ∠LMN are supplementary angles, m∠EFG = (3x + 17)°, and m∠LMN = ((1/2) x − 5)°

Solution :

Since the given angles are supplementary, sum of their angles must be 180 degree.

m∠EFG +  m∠LMN = 180

3x + 17 + (1/2) x − 5 = 180

3x + x/2 + 12 = 180

3x + x/2 = 180 - 12

(6x + x)/2 = 168

7x/2 = 168

x = 168(2/7)

x = 48

Problem 11 :

Find the angle measure if ∠5 is a supplement of ∠6, and m∠5 = 78°. Find m∠6.

Solution :

m∠5 + m∠6 = 180

78 + m∠6 = 180

m∠6 = 180 - 78

m∠6 = 102

Problem 12 :

Find the angle measure if ∠7 is a supplement of ∠8, and m∠7 = 109°. Find m∠8.

Solution :

m∠7 + m∠8 = 180

109 + m∠8 = 180

m∠8 = 180 - 109

m∠8 = 71

Problem 13 :

Find the measure of each angle.

Solution :

m∠QRT + m∠TRS = 180

3x + 5 + 10x - 7 = 180

13x - 2 = 180

13x = 180 + 2

13x = 182

x = 182/13

x = 14

Problem 13 :

The arm of a crossing gate moves 42° from a vertical position. How many more degrees does the arm have to move so that it is horizontal?

a) 42°      b) 138°      c)  48°        d) 90°

Solution :

42 + 138 = 180

To make it as horizontal, we need 138 degree.

Write and solve an algebraic equation to find the measure of each angle based on the given description.

Problem 14 :

The measure of one angle is 3° more than 1/2 the measure of its supplement.

Solution :

Let x be the angle. 

Its supplementary angle is 180 - x

x = 1/2(180 - x) + 3

x = 90 - x/2 + 3

x + x/2 = 93

3x/2 = 93

x = 93(2/3)

x = 31(2)

x = 62

Problem 15 :

Two angles form a linear pair. The measure of one angle is 15° less than 2/3 the measure of the other angle.

Solution :

Let x be the one angle.

The other angle = 2/3 of x - 15

(2x/3) - 15

Since these two angles are linear pair, the sum of the angles will be 180 degree.

x + (2x/3) - 15 = 180

(3x + 2x)/3 = 180 + 15

5x/3 = 195

x = 195(3/5)

= 117

Applying the value of x, we get

= 2(117)/3 - 15

= 78 - 15

= 63

So, the angles are 63 and 117.