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$In /_\DEF,m/_D=(x-9)^(@),m/_E=(3x-6)^(@), and m/_F=(10 x-15)^(@). Find m/_E. Answer:$

In $\triangle \mathrm{DEF}, \mathrm{m} \angle D=(x-9)^{\circ}, \mathrm{m} \angle E=(3 x-6)^{\circ}$ , and $\mathrm{m} \angle F=(10 x-15)^{\circ}$ . Find $\mathrm{m} \angle E$ . $\newline$ Answer:

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Compound interest: word problems

Full solution

Q. In

\triangle \mathrm{DEF}, \mathrm{m} \angle D=(x-9)^{\circ}, \mathrm{m} \angle E=(3 x-6)^{\circ}

, and

\mathrm{m} \angle F=(10 x-15)^{\circ}

. Find

\mathrm{m} \angle E

.

\newline

Answer:

Identify Triangle Angles: Identify the relationship between the angles in a triangle. The sum of the angles in any triangle is always $180$ degrees.
Set Up Equation: Set up the equation based on the given angle expressions. $\newline$ $\frac{m}{\angle D} + \frac{m}{\angle E} + \frac{m}{\angle F} = 180^\circ$ $\newline$ Substitute the given expressions for $\frac{m}{\angle D}$ , $\frac{m}{\angle E}$ , and $\frac{m}{\angle F}$ . $\newline$ $(x - 9)^\circ + (3x - 6)^\circ + (10x - 15)^\circ = 180^\circ$
Combine Like Terms: Combine like terms to simplify the equation. $\newline$ $x - 9 + 3x - 6 + 10x - 15 = 180$ $\newline$ Combine the $x$ terms and the constant terms. $\newline$ $14x - 30 = 180$
Solve for x: Solve for x. $\newline$ Add $30$ to both sides of the equation. $\newline$ $14x - 30 + 30 = 180 + 30$ $\newline$ $14x = 210$ $\newline$ Divide both sides by $14$ . $\newline$ $x = \frac{210}{14}$ $\newline$ $x = 15$
Substitute x Value: Substitute the value of $x$ back into the expression for $m/\angle E$ to find its measure. $\newline$ $m/\angle E = (3x - 6)^\circ$ $\newline$ $m/\angle E = (3(15) - 6)^\circ$ $\newline$ $m/\angle E = (45 - 6)^\circ$ $\newline$ $m/\angle E = 39^\circ$

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