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In /_\CDE, bar(CE) is extended through point E to point F,m/_CDE=(2x+12)^(@), m/_DEF=(7x-6)^(@), and m/_ECD=(2x+3)^(@). Find m/_CDE. Answer:

In △CDE,CE‾ \triangle \mathrm{CDE}, \overline{C E} is extended through point E \mathrm{E} to point F,m∠CDE=(2x+12)∘ \mathrm{F}, \mathrm{m} \angle C D E=(2 x+12)^{\circ} , m∠DEF=(7x−6)∘ \mathrm{m} \angle D E F=(7 x-6)^{\circ} , and m∠ECD=(2x+3)∘ \mathrm{m} \angle E C D=(2 x+3)^{\circ} . Find m∠CDE \mathrm{m} \angle C D E .\newlineAnswer:

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Full solution

Q. In △CDE,CE‾ \triangle \mathrm{CDE}, \overline{C E} is extended through point E \mathrm{E} to point F,m∠CDE=(2x+12)∘ \mathrm{F}, \mathrm{m} \angle C D E=(2 x+12)^{\circ} , m∠DEF=(7x−6)∘ \mathrm{m} \angle D E F=(7 x-6)^{\circ} , and m∠ECD=(2x+3)∘ \mathrm{m} \angle E C D=(2 x+3)^{\circ} . Find m∠CDE \mathrm{m} \angle C D E .\newlineAnswer:
  1. Identify Relationship: Identify the relationship between the angles.\newlineSince CECE is a straight line, the angles CDECDE and DEFDEF form a linear pair and therefore add up to 180180 degrees.
  2. Set Up Equation: Set up the equation based on the linear pair relationship.\newlinem/_CDE+m/_DEF=180∘m/\_CDE + m/\_DEF = 180^\circ\newlineSubstitute the given expressions for m/_CDEm/\_CDE and m/_DEFm/\_DEF.\newline(2x+12)∘+(7x−6)∘=180∘(2x + 12)^\circ + (7x - 6)^\circ = 180^\circ
  3. Combine and Solve: Combine like terms and solve for xx.2x+12+7x−6=1802x + 12 + 7x - 6 = 1809x+6=1809x + 6 = 180
  4. Subtract and Simplify: Subtract 66 from both sides of the equation.\newline9x+6−6=180−69x + 6 - 6 = 180 - 6\newline9x=1749x = 174
  5. Divide for x Value: Divide both sides by 99 to find the value of x.\newline9x9=1749\frac{9x}{9} = \frac{174}{9}\newlinex=19.333…x = 19.333\ldots
  6. Substitute for CDE: Substitute the value of xx back into the expression for m/_CDEm/\_CDE.
    m/_CDE=(2x+12)@m/\_CDE = (2x + 12)^{@}
    m/_CDE=(2(19.333...)+12)@m/\_CDE = (2(19.333...) + 12)^{@}
    m/_CDE=(38.666...+12)@m/\_CDE = (38.666... + 12)^{@}
    m/_CDE=50.666...@m/\_CDE = 50.666...^{@}

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