document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1206533>Question 1206533</A>: <I><SPAN STYLE=\"word-break: break-all;\"> AB is a chord of a circle and the tangents A and B meet at T. C is a point on the minor arc AB. If ∠ATB = 54° and ∠CBT = 23°, calculate ∠CAT. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #844081 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=844081\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "Answer: <font color=red size=4>40 degrees</font>\r<BR>\n" );
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document.write( "Explanation\r<BR>\n" );
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document.write( "This is likely what the diagram looks like.<BR>\n" );
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document.write( "The diagram was made with <a  rel=nofollow HREF=\"https://www.geogebra.org/calculator\">GeoGebra</a>\r<BR>\n" );
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document.write( "Let D be the center of the circle. Draw segments DA and DB which are radii.<BR>\n" );
document.write( "Focus only on quadrilateral DATB.<BR>\n" );
document.write( "For any quadrilateral, the four inside angles must always add to 360 degrees.<BR>\n" );
document.write( "D+A+T+B = 360<BR>\n" );
document.write( "D+90+54+90 = 360<BR>\n" );
document.write( "D+234 = 360<BR>\n" );
document.write( "D = 360-234<BR>\n" );
document.write( "D = 126<BR>\n" );
document.write( "This is the measure of angle ADB. It is also the measure of minor arc AB. Central angles subtending an arc will have the same measure.<BR>\n" );
document.write( "Note that angles A and B are 90 degrees each since they are at points of tangency.\r<BR>\n" );
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document.write( "Let E be some point on the circle that's not on minor arc AB. Refer to the diagram below.<BR>\n" );
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document.write( "Major arc AEB and minor arc AB glue together to form the entire circle. There are no gaps and no overlaps between the arcs. The arcs partition the circle's circumference.\r<BR>\n" );
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document.write( "Because of this we know that:<BR>\n" );
document.write( "(major arc AEB) + (minor arc AB) = 360<BR>\n" );
document.write( "(major arc AEB) + 126 = 360<BR>\n" );
document.write( "major arc AEB = 360 - 126<BR>\n" );
document.write( "major arc AEB = 234\r<BR>\n" );
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document.write( "Now we'll use the inscribed angle theorem to determine inscribed angle ACB.<BR>\n" );
document.write( "angle ACB = (1/2)*(major arc AEB)<BR>\n" );
document.write( "angle ACB = (1/2)*(234)<BR>\n" );
document.write( "angle ACB = 117<BR>\n" );
document.write( "We'll use this later after a slight detour in the next section.\r<BR>\n" );
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document.write( "-------------------------------\r<BR>\n" );
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document.write( "Draw a segment connecting points D and T.<BR>\n" );
document.write( "Quadrilateral DATB is split into triangle DAT and triangle DBT.<BR>\n" );
document.write( "They are both right triangles due to the points of tangency.<BR>\n" );
document.write( "We can use the hypotenuse leg (HL) theorem to prove those right triangles are congruent, and it consequently means AT = TB.\r<BR>\n" );
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document.write( "Since AT = TB, we determine that triangle BAT is isosceles.\r<BR>\n" );
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document.write( "Furthermore, it means angle ABT = angle BAT since they are the base angles. The congruent base angles are opposite the congruent sides.\r<BR>\n" );
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document.write( "Focus only on triangle ATB.<BR>\n" );
document.write( "The three inside angles must add to 180 degrees.<BR>\n" );
document.write( "A+T+B = 180<BR>\n" );
document.write( "x+54+x = 180<BR>\n" );
document.write( "2x+54 = 180<BR>\n" );
document.write( "2x = 180-54<BR>\n" );
document.write( "2x = 126<BR>\n" );
document.write( "x = 126/2<BR>\n" );
document.write( "x = 63\r<BR>\n" );
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document.write( "Base angles ABT and BAT are 63 degrees each.\r<BR>\n" );
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document.write( "Then,<BR>\n" );
document.write( "(angle CBA) + (angle CBT) = angle ABT<BR>\n" );
document.write( "angle CBA = (angle ABT) - (angle CBT)<BR>\n" );
document.write( "angle CBA = (63) - (23)<BR>\n" );
document.write( "angle CBA = 40\r<BR>\n" );
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document.write( "The conclusions of the first two sections are<BR>\n" );
document.write( "angle ACB = 117<BR>\n" );
document.write( "angle CBA = 40<BR>\n" );
document.write( "These represent angles C and B in triangle ABC. \r<BR>\n" );
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document.write( "Focus only on triangle ABC.<BR>\n" );
document.write( "A+B+C = 180<BR>\n" );
document.write( "A+117+40 = 180<BR>\n" );
document.write( "A+157 = 180<BR>\n" );
document.write( "A = 180-157<BR>\n" );
document.write( "A = 23<BR>\n" );
document.write( "This is the measure of angle CAB.\r<BR>\n" );
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document.write( "Then we have one last set of steps.<BR>\n" );
document.write( "(angle CAB)+(angle CAT) = angle BAT<BR>\n" );
document.write( "(23)+(angle CAT) = 63<BR>\n" );
document.write( "angle CAT = 63-23<BR>\n" );
document.write( "<font color=red>angle CAT = 40</font> is the final answer.\r<BR>\n" );
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document.write( "My response is a bit long-winded. Another tutor may provide a much more efficient pathway.<BR>\n" );
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