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if ABC is an isosceles triangle and DBE is an equilateral triangle find each missing measure

if ABC is an isosceles triangle and DBE is an equilateral triangle find each missing measure class=

Respuesta :

Answer:

∠1=∠9=43°,  ∠2=∠7= 17°,  ∠4=∠5=∠6=60°, ∠3=∠8=120°

Step-by-step explanation:

Given ABC is an isosceles triangle and DBE is an equilateral triangle. we have to find each missing measure.

In triangle ABC,

∠1=∠9    (∵isosceles triangle)

⇒ 4x+3=9x-47

⇒ 9x-4x=3+47 ⇒ 5x=50 ⇒ x=10

Hence, ∠9=∠1=(4x+3)°=4(10)+3=43°

Also, given ΔBDE is an equilateral triangle, and all angle of equilateral triangle are equal.

∴ ∠4+∠5+∠6=180°

⇒ ∠4+∠4+∠4=180° ⇒ 3∠4 = 180° ⇒ ∠4 = 60°

∴ ∠4=∠5=∠6=60°

By exterior angle property, ∠3=∠5+∠6=60°+60°=120°

                                             ∠8=∠5+∠4=60°+60°=120°

In ΔABD, ∠1+∠2+∠3=180°

             ⇒ 43°+120°+∠2=180°⇒ ∠2 = 17°

In ΔABD, ∠9+∠8+∠7=180°

             ⇒ 43°+120°+∠7=180°⇒ ∠7= 17°





           

MrRoyal

The base angles of an isosceles triangle are equal.

The measure of the angles are:

  • [tex]\mathbf{\angle 1 = 43}[/tex]      [tex]\mathbf{\angle 2=17}[/tex]        [tex]\mathbf{\angle 3=120}[/tex]
  • [tex]\mathbf{\angle 4=\angle 5=\angle 6=60}[/tex]
  • [tex]\mathbf{\angle 7 = 17}[/tex]        [tex]\mathbf{\angle 8=120}[/tex]        [tex]\mathbf{\angle 9 = 43}[/tex]

In ABC, we have:

[tex]\mathbf{\angle 1 = \angle 9}[/tex] --- base angles of an isosceles triangle

This gives

[tex]\mathbf{4x+3=9x-47}[/tex]

Collect like terms

[tex]\mathbf{9x - 4x= 3+47}[/tex]

[tex]\mathbf{5x= 50}[/tex]

Divide both sides by 5

[tex]\mathbf{x= 10}[/tex]

So, we have:

[tex]\mathbf{\angle 1 = 4x + 3}[/tex]

[tex]\mathbf{\angle 1 = 4 \times 10 + 3}[/tex]

[tex]\mathbf{\angle 1 = 43}[/tex]

Also:

[tex]\mathbf{\angle 9 = 43}[/tex]

Angles in an equilateral triangle are equal.

So:

[tex]\mathbf{\angle 4=\angle 5=\angle 6=60}[/tex]

In triangle ABC, we have:

[tex]\mathbf{\angle 1 + \angle 2+\angle 5+\angle 7 + \angle 9 =180}[/tex] -- angles in a triangle

[tex]\mathbf{43 + \angle 2+\angle 5+\angle 7 + 43=180}[/tex]

Collect like terms

[tex]\mathbf{\angle 2+\angle 5+\angle 7 =180 - 43 - 43}[/tex]

[tex]\mathbf{\angle 2+\angle 5+\angle 7 =94}[/tex]

Substitute 60 for <5

[tex]\mathbf{\angle 2+60+\angle 7 =94}[/tex]

[tex]\mathbf{\angle 2+\angle 7 =94 - 60}[/tex]

[tex]\mathbf{\angle 2+\angle 7 =34}[/tex]

 

 By exterior angle property, we have:

[tex]\mathbf{\angle 3=\angle 5+\angle 6}[/tex]

So, we have:

[tex]\mathbf{\angle 3=60+60}[/tex]

[tex]\mathbf{\angle 3=120}[/tex]

Similarly

[tex]\mathbf{\angle 8=\angle 4+\angle 5}[/tex]

[tex]\mathbf{\angle 8=\angle 60+\angle 60}[/tex]

[tex]\mathbf{\angle 8=120}[/tex]

In ΔABD, we have:

[tex]\mathbf{\angle 1+\angle 2+\angle 3=180}[/tex]

Substitute known values

[tex]\mathbf{43+\angle 2+120=180}[/tex]

Collect like terms

[tex]\mathbf{\angle 2=180 - 120 - 43}[/tex]

[tex]\mathbf{\angle 2=17}[/tex]

Recall that:

[tex]\mathbf{\angle 2+\angle 7 =34}[/tex]

So, we have:

[tex]\mathbf{17 + \angle 7 = 34}[/tex]

Subtract 17 from both sides

[tex]\mathbf{\angle 7 = 17}[/tex]

Read more about isosceles and equilateral triangles at:

https://brainly.com/question/6238271

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