Question 1169135: Points A, B, C, and D are arranged on a circle so that arc AB=27, arc BC=30,
arc CD=44, and arc DA=259. Let Q be the intersection of the Chords AC and BD.
Find angles CAD and AQD.
Found 2 solutions by Edwin McCravy, greenestamps:
Answer by Edwin McCravy(20064)   (Show Source):
You can put this solution on YOUR website! Points A, B, C, and D are arranged on a circle so that arc AB=27, arc BC=30,
arc CD=44, and arc DA=259. Let Q be the intersection of the Chords AC and
BD. Find angles CAD and AQD.
Since the sum or arcs AD+BC+CD+DA = 27+30+44+259=360,
we can consider them as arc measures in degrees.
Angle CAD is an inscribed angle, so its measure is
1/2 the measure of its intercepted arc's measure. It
intercepts arc CD, which is 44°.
So angle CAD has measure which is 1/2 of that or 22°.
Angle BDA is also an inscribed angle, so its measure is
1/2 the measure of its intercepted arc's measure. It
intercepts arc AB, which is 27°.
So angle BDA has measure which is 1/2 of that or 13.5°.
Since the sum of the interior angles of triangle AQD is
180°, then angle AQD has measure 180°-22°-13.5° = 144.5°.
Edwin
Answer by greenestamps(13209)  (Show Source):
You can put this solution on YOUR website!
Here is an alternative approach to one part of the problem, compared to the response from the other tutor.
As in the other response, Angle CAD is an inscribed angle that cuts off arc CD, which measures 44 degrees. The measure of the inscribe angle is half the measure of the arc.
ANSWER: Angle CAD is 44/2 = 22 degrees.
Angle BQD is formed by the intersection of two chords. The measure of the angle is half the sum of the measures of the two intercepted arcs.
Those arc measures are 30 degrees and 259 degrees.
ANSWER: The measure of angle BQD is (259+30)/2 = 289/2 = 144.5 degrees.
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