Question

triangle abc is inscribed in a circle centered at point o.
order the measures of the three intercepted arcs from least to greatest.
\\(\boldsymbol{\mathrm{m}\\ \overarc{bc}}\\) \\(\boldsymbol{\mathrm{m}\\ \overarc{ab}}\\) \\(\boldsymbol{\mathrm{m}\\ \overarc{ac}}\\)
\\(\square < \square < \square\\)

Explanation:

Step1: Find $m\overset{\frown}{AB}$

An inscribed angle is half its arc.
$m\overset{\frown}{AB} = 2\times 63^\circ = 126^\circ$

Step2: Find $m\overset{\frown}{AC}$

Total circle is $360^\circ$.
$m\overset{\frown}{AC} = 360^\circ - 118^\circ - 126^\circ = 116^\circ$

Step3: Order the arcs

Compare arc measures: $116^\circ < 118^\circ < 126^\circ$, so $m\overset{\frown}{AC} < m\overset{\frown}{BC} < m\overset{\frown}{AB}$ was a mistake, correction:

Step1: Correct inscribed angle rule

Inscribed angle $\angle ACB$ intercepts $\overset{\frown}{AB}$
$m\overset{\frown}{AB} = 2\times m\angle ACB = 2\times63^\circ=126^\circ$

Step2: Calculate remaining arc sum

Total circle $360^\circ$, subtract known arcs:
$m\overset{\frown}{AC}=360^\circ - m\overset{\frown}{BC} - m\overset{\frown}{AB}=360^\circ-118^\circ-126^\circ=116^\circ$

Step3: Order from least to greatest

Compare $116^\circ, 118^\circ, 126^\circ$:
$m\overset{\frown}{AC} < m\overset{\frown}{BC} < m\overset{\frown}{AB}$

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Corrected Final Answer:

$m\overset{\frown}{AC} < m\overset{\frown}{BC} < m\overset{\frown}{AB}$

Step1: Find $\boldsymbol{m\overset{\frown}{AB}}$

Inscribed angle = ½ its arc.
$m\overset{\frown}{AB} = 2\times 63^\circ = 126^\circ$

Step2: Find $\boldsymbol{m\overset{\frown}{AC}}$

Total circle is $360^\circ$.
$m\overset{\frown}{AC} = 360^\circ - 118^\circ - 126^\circ = 116^\circ$

Step3: Order the three arcs

Compare the three arc measures: $116^\circ < 118^\circ < 126^\circ$, so $m\overset{\frown}{AC} < m\overset{\frown}{BC} < m\overset{\frown}{AB}$

Answer:

$m\overset{\frown}{AB} < m\overset{\frown}{BC} < m\overset{\frown}{AC}$