Question

5. Name the angle that is supplementary to COD.

A
F
OZCOA
O ZBOC
OLCOB
O ZBOA
B
O
۶,
D
(1 point)

Answer 1

Example 2:

Find m∠P  and  m∠Q

if ∠P  and  ∠Q

are supplementary, m∠P=2x+15

, and m∠Q=5x−38

.

The sum of the measures of two supplementary angles is 180°

.

So, m∠P+m∠Q=180°

Substitute 2x+15

for m∠P

and 5x−38

for m∠Q

.

2x+15+5x−38=180°

Combine the like terms. We get:

7x−23=180°

Add 23

to both the sides. We get:

7x=203°

Divide both the sides by 7

.

7x7=203°7

Simplify.

x=29°

To find m∠P

, substitute 29

for x

in 2x+15

.

2(29)+15=58+15

Simplify.

58+15=73

So, m∠P=73°

.

To find m∠Q

, substitute 29

for x

in 5x−38

.

5(29)−38=145−38

Simplify.

145−38=107

So, m∠Q=107°

.Example 2:

Find m∠P  and  m∠Q

if ∠P  and  ∠Q

are supplementary, m∠P=2x+15

, and m∠Q=5x−38

.

The sum of the measures of two supplementary angles is 180°

.

So, m∠P+m∠Q=180°

Substitute 2x+15

for m∠P

and 5x−38

for m∠Q

.

2x+15+5x−38=180°

Combine the like terms. We get:

7x−23=180°

Add 23

to both the sides. We get:

7x=203°

Divide both the sides by 7

.

7x7=203°7

Simplify.

x=29°

To find m∠P

, substitute 29

for x

in 2x+15

.

2(29)+15=58+15

Simplify.

58+15=73

So, m∠P=73°

.

To find m∠Q

, substitute 29

for x

in 5x−38

.

5(29)−38=145−38

Simplify.

145−38=107

So, m∠Q=107°

.