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$In /_\PQR, bar(PR) is extended through point R to point S,m/_RPQ=(3x+15)^(@), m/_QRS=(9x-13)^(@), and m/_PQR=(2x-4)^(@). What is the value of x? Answer:$

In $\triangle \mathrm{PQR}, \overline{P R}$ is extended through point $\mathrm{R}$ to point $\mathrm{S}, \mathrm{m} \angle R P Q=(3 x+15)^{\circ}$ , $\mathrm{m} \angle Q R S=(9 x-13)^{\circ}$ , and $\mathrm{m} \angle P Q R=(2 x-4)^{\circ}$ . What is the value of $x ?$ $\newline$ Answer:

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Math Problems

Algebra 2

Csc, sec, and cot of special angles

Full solution

Q. In

\triangle \mathrm{PQR}, \overline{P R}

is extended through point

\mathrm{R}

to point

\mathrm{S}, \mathrm{m} \angle R P Q=(3 x+15)^{\circ}

\mathrm{m} \angle Q R S=(9 x-13)^{\circ}

, and

\mathrm{m} \angle P Q R=(2 x-4)^{\circ}

. What is the value of

x ?

\newline

Answer:

Interior Angles of Triangle: The sum of the interior angles of a triangle is always $180$ degrees. Since $PR$ is extended through point $R$ to point $S$ , angle $QRS$ is an exterior angle to triangle $PQR$ . According to the exterior angle theorem, the measure of an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles.
Exterior Angle Theorem Equation: Let's write the equation based on the exterior angle theorem: $\newline$ $m/_{\text{RPQ}} + m/_{\text{PQR}} = m/_{\text{QRS}}$ $\newline$ $(3x + 15) + (2x - 4) = (9x - 13)$
Combine Like Terms: Now, let's combine like terms: $\newline$ $3x + 2x + 15 - 4 = 9x - 13$ $\newline$ $5x + 11 = 9x - 13$
Solve for x: Next, we will solve for $x$ by moving the $x$ terms to one side and the constant terms to the other side: $5x - 9x = -13 - 11$ $-4x = -24$
Final Solution: Now, divide both sides by $-4$ to solve for $x$ : $x = \frac{-24}{-4}$ $x = 6$