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EXAMPLE 3 - PowerPoint PPT Presentation

o. ALGEBRA Given that m LKN =145 , find m LKM and m MKN. STEP 1. Write and solve an equation to find the value of x. m LKN = m LKM + m MKN. o. o. o. 145 = (2 x + 10) + (4 x – 3). EXAMPLE 3. Find angle measures. SOLUTION. Angle Addition Postulate.

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ALGEBRAGiven that m LKN =145 , find m LKM andm MKN.

STEP 1

Write and solve an equation to find the value of x.

mLKN = m LKM + mMKN

o

o

o

145 = (2x + 10)+ (4x – 3)

EXAMPLE 3

Find angle measures

SOLUTION

Substitute angle measures.

145 = 6x + 7

Combine like terms.

138 = 6x

Subtract 7 from each side.

23 = x

Divide each side by 6.

STEP 2

Evaluate the given expressions when x = 23.

mLKM = (2x+ 10)° = (2 23+ 10)° = 56°

mMKN = (4x– 3)° = (4 23– 3)° = 89°

So, m LKM = 56°and m MKN = 89°.

EXAMPLE 3

Find angle measures

3. Given that KLMis straight angle, find mKLN andm NLM.

STEP 1

Write and solve an equation to find the value of x.

m KLM + m NLM

= 180°

(10x – 5)° + (4x +3)°

= 180°

= 180

14x – 2

= 182

14x

x

= 13

for Example 3

GUIDED PRACTICE

Find the indicated angle measures.

SOLUTION

Straight angle

Substitute angle measures.

Combine like terms.

Subtract 2 from each side.

Divide each side by 14.

STEP 2

Evaluate the given expressions when x = 13.

mKLM = (10x– 5)° = (10 13– 5)° = 125°

mNLM = (4x+ 3)° = (4 13+ 3)° = 55°

mKLM

= 125°

mNLM

= 55°

for Example 3

GUIDED PRACTICE

STEP 1

Write and solve an equation to find the value of x.

m EFG + m HFG

m EFG

EFG is a right angle

=

= 90°

(2x + 2)° + (x +1)°

= 90°

= 90

3x + 3

= 87

3x

x

= 29

for Example 3

GUIDED PRACTICE

4. Given that EFGis a right angle, find mEFH andm HFG.

SOLUTION

Substitute angle measures.

Combine like terms.

Subtract 3 from each side.

Divide each side by 3.

STEP 2

Evaluate the given expressions when x = 29.

mEFH = (2x+ 2)° = (2 29 +2)° = 60°

mHFG = (x+ 1)° = (29+ 1)° = 30°

mEFG

= 60°

mHFG

= 30°