In trapezoid ABCD, BC || AD, angle ABD =105°, angle A=43°, and angle C=141°. Find angle CBD, in degrees.
Hi Mellie,
In trapezoid ABCD, BC || AD, angle ABD =105°, angle A=43°, and angle C=141°. Find angle CBD, in degrees.
ok, the first thing I did is tried to work out what the trapezium looked like.
THEN the answer became obvious.
BUT
it also became ovvious that I did not need to know what the trapezium looked like in ordet to answer this question.
Draw ANY trapezium where BC||AD
Label <ABD as 105 degrees. (It doesn't need to LOOK anythink like 105 degrees, mine looks like an acute angle)
Label <A as 43 degrees
Label < C as 141 degrees
Now you can see that <CBD is an alternate angle on parallel lines to <BDA. So they are congruent.
< BDA+105+43=180 Angle sum of triangle ABD
<BDA = 32 degrees
Therefore <CBD = 32 degrees
Here is a diagram most definitely not to scale.
ABD = 105° and ABD will form a triangle.....then angle ADB will = 180 - 43 -105 = 32°
And, since AD is parallel to BC, angle CBD will have the same measure as angle ABD = 32° because they are alternating interior angles between these parallels