+0

0
326
1

Points $A$ and $B$ lie on a circle centered at $O$, and point $X$ is outside the circle such that $\overline{AX}$ and $\overline{BX}$ are tangent to the circle. If $\angle AXO = 26^{\circ}$, then what is the measure of minor arc $AB$, in degrees?

Feb 9, 2020

#1
+21955
0

Points  A  and  B  divide the circle into two arcs:  major arc(AB)  and  minor arc(AB).

The size of angle(AXO)  =  26°.

This means that the size of angle(BXO) is also 26°.

The total size of angle(AXB)  =  52°.

Since we want to find the size of minor arc(AB), call minor arc(AB) = x.

Since the sum of the major arc(AB) and minor arc(AB) = 360°, major arc(AB)  =  360° - x.

The appropriate formula is:  [ major arc(AB) - minor arc(AB) ] / 2  =  angle(AXB).

[  (360° - x) - x ] / 2  =  52°

[ 360° - 2x] / 2  =  52°

360° - 2x  =  104°

256°  =  2x

x  =  128°

Feb 9, 2020
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}