In trapezoid ABCD, BC || AD, angle ABD =105°, angle A=43°, and angle C=141°. Find angle CBD, in degrees.

Mellie
Dec 12, 2015

#1**+15 **

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.

Melody
Dec 12, 2015

#2**+10 **

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

CPhill
Dec 12, 2015

#3