D is on side AB of Triangle ABC. We know triangle ABC\(\sim\)triangle ACD and angle A=48. If BC=BD, what is angle B in degrees?
here is the image:
https://docs.google.com/document/d/1f9ygPhFBH44Ko1mmiGzzfVsAWfiOA3pX50IsnYQ9OBM/edit?usp=sharing
I'll add the pic - Melody
I get 28 degrees. (which I have not checked)
Let's see what catmg comes up with.
Nice. :))
I really miss going on this forum more often.
Sadly, school is quite busy.
=^._.^=
You are the one who did the nice work Catmg.
Yes school is busy, so is life for younger people.
Most of the people who are here a lot, to answer questions I mean, are retired.
It can continue to be a nice hobby for you though :))
Hello CCeed :))
Let angle B = x
angle A = 48
angle C = 132 - x
Since ABC ~ ACD
angle B = angle ACD = x
angle C = angle CDA = 132 - x
Since BC = BD
BCD = BDC
BCD + BDC + B = 180 degrees
BDC + BDC + x = 180 degrees
BDC = (180-x)/2
CDA + CDB = 180
132 - x + (180-x)/2 = 180
x = 28
I hope this helped. :))
=^._.^=
I find it quite difficult to get my head around geometry questions such as this.
So I will give you some insight as to how I go about it.
I drew the diagram and shade the smaller one.
then I said the smallest side of the big triangle is AC which corresponds to the smallest side of the small diagram AD
then I said the biggest side of the big triangle is AB which corresponds to the biggest side of the small diagram AC
Now I draw the corresponding triangles like this:
And then I take it from there.
I suppose it is too slow long term but while you are getting the hang of it it can help.