Let us have a triangle ABC and a point D on BC such that BD = DC = DA. If angle ABC = 50 degrees, then how many degrees are in angle ACB?

Please help and explain ASAP, thank you!

Otterstar Jul 31, 2020

#1**+1 **

Hewo, Otterstar!~

So we have a triangle, ABC, where Point D is on BC, such that BD=DC=DA. (I had to draw it like seven times to get the scale right (I don't know why I bothered to draw it so perfectly)). Now, we're told that ABC is 50.

So, since DA=DC=BD, I'm going ahead and assuming that DA perfectly chops angle ABC in half, meaning, Angle DAC= 50/2=25. (Also BAD but we don't care about that.)

Great. Now what? We still need one more angle to find ACD.

So we go back to the part where ABC is perfectly chopped down in half. That means that angle ADC is 90 degrees, (also ADB but we don't care about that), and so, we have two angles of a triangle, and we need to find the last one...

DAC+ACD+ADC=180

180=DAC+ACD+ADC

180=ACD+90+25

180=ACD+115

65=ACD

(ACD=ACB( Just different ways of saying it))

Therefore, ACD=65, and by extension, ACB=65.

Reply to my answer if I made a mistake, or have any questions!~

-TheLovely1

TheLovely1 Jul 31, 2020