Find Length of Arc EF in Triangle ABC

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SUMMARY

The length of arc EF in triangle ABC, where angle A is 70 degrees and point O is the midpoint of segment BC measuring 12 units, is calculated using the properties of circles and triangles. The angle BFC is a right angle, making triangle ABF a right-angled triangle with angle ABF measuring 20 degrees. Consequently, angle EOF at the center is 40 degrees. Given that the radius of circle O is 6 units, the length of arc EF is determined to be approximately 4.1888 units using the formula for arc length.

PREREQUISITES
  • Understanding of basic triangle properties and angles
  • Knowledge of circle geometry, specifically arc length calculations
  • Familiarity with right-angled triangles and their properties
  • Ability to apply the relationship between angles at the center and circumference
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  • Study the properties of circles and their arcs in geometry
  • Learn about the relationship between angles in triangles and circles
  • Explore advanced triangle theorems, such as the Inscribed Angle Theorem
  • Practice calculating arc lengths with varying angles and radii
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Students studying geometry, mathematics educators, and anyone interested in solving problems related to triangles and circles.

Albert1
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$\triangle ABC$,with $\angle A=70^o,$point $O$ is the midpoint of segment $\overline{BC}=12$
circle $O$(wth center $O$ and radius $\overline{BO})$, meets with $\overline{AB},\overline{AC}$ at points $E$ and $F$ respectively please find the length of arc $EF$
 
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Albert said:
$\triangle ABC$,with $\angle A=70^o,$point $O$ is the midpoint of segment $\overline{BC}=12$
circle $O$(wth center $O$ and radius $\overline{BO})$, meets with $\overline{AB},\overline{AC}$ at points $E$ and $F$ respectively please find the length of arc $EF$
[sp]
In the diagram, the angle $BFC$ is a right angle (angle in a semicircle). So $ABF$ is a right-angled triangle and $\angle ABF = 20^\circ$. Therefore $\angle EOF = 40^\circ$ (angle at centre = twice angle at circumference). If $BC$ is 12 units then the radius of the circle is 6 units, and the arc $EF$ is $2\pi\cdot 6\cdot\frac{40}{360} = \frac43\pi \approx 4.1888$ units.

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perfect ! very good solution !
 

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