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In ΔCDE, \overline{CE}
CE
is extended through point E to point F, m∠DEF = (4x+5)^{\circ}(4x+5)
∘
, m∠ECD = (x+9)^{\circ}(x+9)
∘
, and m∠CDE = (x+8)^{\circ}(x+8)
∘
. Find m∠CDE.

Respuesta :

calculista

Answer:

[tex]m\angle CDE=14^o[/tex]

Step-by-step explanation:

we know that

An exterior angle of a triangle is equal to the sum of the opposite interior angles.

see the attached figure to better understand the problem

so

∠DEF=∠ECD+∠CDE

substitute the given values

[tex](4x+5)^o=(x+9)^o+(x+8)^o[/tex]

solve for x

Combine like terms

[tex](4x+5)^o=(2x+17)^o[/tex]

Group terms

[tex]4x-2x=17-5[/tex]

[tex]2x=12[/tex]

[tex]x=6[/tex]

Find the measure of angle CDE

[tex]m\angle CDE=(x+8)^o[/tex]

substitute the value of x

[tex]m\angle CDE=(6+8)=14^o[/tex]

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