geochallenge 3

find the measure of $\angle gcf$.

given:
$d$ is the incenter.
$\angle gcd = (7x - 1)^\circ$
$\angle fcd = (5x + 5)^\circ$

Question

geochallenge 3

find the measure of $\angle gcf$.

given:
$d$ is the incenter.
$\angle gcd = (7x - 1)^\circ$
$\angle fcd = (5x + 5)^\circ$

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Incenter Theorem", "Angle Bisector Theorem" ], "new_concepts": [ "Incenter Angle Bisector Application" ], "current_concepts": [ "Incenter Theorem", "Angle Bisector Theorem", "Incenter Angle Bisector Application" ] } ## Identify the geometric properties of the incenter Using the *Incenter Theorem* and *Angle Bisector Theorem* knowledge points $$ \begin{aligned} &\text{Since } D \text{ is the incenter of } \triangle ABC,\ &\text{the segment } CD \text{ is the angle bisector of } \angle GCF.\ &\text{Therefore, } \angle GCD = \angle fCD. \end{aligned} $$ ## Set up the algebraic equation Using the *Incenter Theorem* knowledge point $$ \begin{aligned} &7x - 1 = 5x + 5 \end{aligned} $$ ## Solve for x Using the *Incenter Theorem* knowledge point $$ \begin{aligned} &2x = 6\ &x = 3 \end{aligned} $$ ## Calculate the measure of each half-angle Using the *Incenter Theorem* knowledge point $$ \begin{aligned} &\angle GCD = 7(3) - 1 = 20^\circ\ &\angle fCD = 5(3) + 5 = 20^\circ \end{aligned} $$ ## Find the total measure of angle GCF Using the *Incenter Theorem* knowledge point $$ \begin{aligned} &\angle GCF = \angle GCD + \angle fCD\ &\angle GCF = 20^\circ + 20^\circ = 40^\circ \end{aligned} $$ $40^\circ$ { "subject": "Mathematics", "question_type": "Free Response", "knowledge_point": [ "Mathematics", "Geometry", "Incenter Theorem" ] }

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QUESTION IMAGE

Question

Explanation:

Identify the geometric properties of the incenter

Set up the algebraic equation

Solve for x

Calculate the measure of each half-angle

Find the total measure of angle GCF

Answer: