Ok, here's the data I'm given: In the figure below, A, D, B, and G are collinear. Angle CAD measures 76 degrees, BCD = 47 degrees and CBG = 140 degrees. What is the degree measure of angle ACD?
2007-03-11 09:55:58 · 4 answers · asked by eahummel82 3 in Science & Mathematics ➔ Mathematics
140 = 76 +47 + angle ACD
So angle ACD = 140-76-47 = 17 degrees
2007-03-11 10:18:08 · answer #1 · answered by ironduke8159 7 · 0⤊ 0⤋
angle 140= cbg =180-cbd
cbd =40 degrees
bcd forms a triange with two know angles:
bdc = 180 -40-47 =93 degrees
bdc + adc must equal 180 since a d and b are collinear
adc = 180 -93 = 87
adc form a triangle with two known angles:
cad + adc + acd = 180
76 + 87 + acd = 180
acd = 17 degrees
2007-03-11 10:08:23 · answer #2 · answered by bignose68 4 · 0⤊ 0⤋
angle CDB = 140 - 47 = 93 degrees
angle ACD = 93 - 76 = 17 degrees
Note: the measure of an exterior angle is equal to the sum of the measures of the two remote interior angles.
2007-03-11 10:01:55 · answer #3 · answered by Newbody 4 · 0⤊ 0⤋
mâ DCB = 140-47 = 93⁰, exterior angle theorem of a triangle
mâ ACD = 93-76 = 17⁰, exterior angle theorem of a triangle.
2007-03-11 10:05:19 · answer #4 · answered by sahsjing 7 · 0⤊ 0⤋