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If m<ECD is six less than five times m<BCE. and m<BCD = 162°, find each measure.(Sorry bout using the wrong signs, but < is meant to represent angle)​

If mltECD is six less than five times mltBCE and mltBCD 162 find each measureSorry bout using the wrong signs but lt is meant to represent angle class=

Respuesta :

Ashraf82

Answer:

m∠BCE = 28° and m∠ECD = 134°

Step-by-step explanation:

* Lets explain how to solve the problem

- The figure has three angles: ∠BCE , ∠ECD , and ∠BCD

- m∠ECD is six less than five times m∠BCE

- That means when we multiply measure of angle BCE by five and

 then subtract six from this product the answer will be the measure

 of angle ECD

∴ m∠ECD = 5 m∠BCE - 6 ⇒ (1)

∵ m∠BCD = m∠BCE + m∠ECD

∵ m∠BCD = 162°

∴ m∠BCE + m∠ECD = 162 ⇒ (2)

- Substitute equation (1) in equation (2) to replace angle ECD by

  angle BCE

∴ m∠BCE + (5 m∠BCE - 6) = 162

- Add the like terms

∴ 6 m∠BCE - 6 = 162

- Add 6 to both sides

∴ 6 m∠BCE = 168

- Divide both sides by 6

∴ m∠BCE = 28°

- Substitute the measure of angle BCE in equation (1) to find the

  measure of angle ECD

∵ m∠ECD = 5 m∠BCE - 6

∵ m∠BCE = 28°

∴ m∠ECD = 5(28) - 6 = 140 - 6 = 134°

* m∠BCE = 28° and m∠ECD = 134°

Answer:

m∠BCE = 28°

m∠ECD = 134°

Step-by-step explanation:

Statement in the question says " m∠ECD is six less than five times m∠BCE.

m∠ECD = 5(m∠BCE) - 6

m∠BCD = 162°

SInce m∠BCD = m∠BCE + m∠ECD

162° = m∠BCE + 5(m∠BCE) - 6

162° = 6(m∠BCE) - 6

162 + 6 = 6(m∠BCE)

168 = 6(m∠BCE)

m∠BCE = [tex]\frac{168}{6}[/tex]

m∠BCE = 28°

Since m∠BCD = m∠BCE + m∠ECD

162 = 28 + m∠ECD

m∠ECD = 162 - 28 = 134°

Therefore, m∠BCE = 28° and m∠ECD = 134° is the answer.

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