Respuesta :
[tex]a. $ x = 23^{\circ}\\b. $ m \angle ECH =28^{\circ}\\c. $ m\angle HCD = 62^{\circ}\\d. $ m \angle GCF = 28^{\circ}\\e. $ m \angle ECG = 152^{\circ}\\f. $ m \angle GCD = 118^{\circ}[/tex]
Given:
[tex]m \angle ECH = x + 5\\m \angle HCD = 3x - 7[/tex]
Note the following:
Since CD is perpendicular to EF, therefore:
[tex]m \angle ECD = 90^{\circ}\\m \angle FCD = 90^{\circ}[/tex]
Thus:
a. Find x
[tex]m \angle ECH + m \angle HCD = m \angle ECD[/tex]
Substitute
[tex](x + 5) + (3x - 7) = 90[/tex]
Add like terms
[tex]x + 5 +3x - 7 = 90\\4x -2 = 90\\4x = 90 + 2\\4x = 92[/tex]
Divide both sides by 4
[tex]x = 23[/tex]
b. Find [tex]m \angle ECH[/tex]
[tex]m \angle ECH = x + 5[/tex]
Plug in the value of x
[tex]m \angle ECH = 23+ 5\\m \angle ECH =28^{\circ}[/tex]
c. Find [tex]m \angle HCD[/tex]
[tex]m \angle HCD = 3x - 7\\m \angle HCD = 3(23) - 7\\m \angle HCD = 62^{\circ}[/tex]
d. Find [tex]m \angle GCF[/tex]
[tex]m \angle GCF = m \angle ECH[/tex] (vertical angles are congruent)
Substitute
[tex]m \angle GCF = 28^{\circ}[/tex]
e. Find [tex]m \angle ECG[/tex]
[tex]m \angle ECG = 180 - m \angle GCF[/tex] (angles on a straight line)
Substitute
[tex]m \angle ECG = 180 - 28\\m \angle ECG = 152^{\circ}[/tex]
f. Find [tex]m \angle GCD[/tex]
[tex]m \angle GCD = m\angle FCD + m \angle GCF[/tex]
Substitute
[tex]m \angle GCD = 90 + 28\\m \angle GCD = 118^{\circ}[/tex]
Therefore:
[tex]a. $ x = 23^{\circ}\\b. $ m \angle ECH =28^{\circ}\\c. $ m\angle HCD = 62^{\circ}\\d. $ m \angle GCF = 28^{\circ}\\e. $ m \angle ECG = 152^{\circ}\\f. $ m \angle GCD = 118^{\circ}[/tex]
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