Given: Triangles ABC and BCD are isosceles, m∠BDC = 30°, and m∠ABD = 155°.

Find m∠ABC, m∠BAC, and m∠DBC.

Question

Given: Triangles ABC and BCD are isosceles, m∠BDC = 30°, and m∠ABD = 155°. 

Find m∠ABC, m∠BAC, and m∠DBC.

nikato · Answer

Do u have a picture?

anonymous · Answer

nikato · Answer

I think the easiest to find first is DBC

nikato · Answer

Isosceles triangle theorem will tell us DBC=DCB.
Also we know a triangle is 180

nikato · Answer

So 
30+2x=180

Solving for x will give u <DBC

anonymous · Answer

75?

anonymous · Answer

Can someone help me for the last two please

anonymous · Answer

$\angle ABD=\angle ABC+\angle DBC$. You are given $\angle ABD$ and $\angle DBC$. Substitution and solving, you will get $\angle ABC$.

anonymous · Answer

Can you help me solve? Im not very good at this stuff

anonymous · Answer

Since $\angle ABD=155^\text{o}$ and $\angle DBC=75^\text{o}$, then from the previous equation, $$155^\text{o}=\angle ABC+75^\text{o}$$

anonymous · Answer

I got 230?

anonymous · Answer

Nope. Subtract 75 both sides.

anonymous · Answer

80?

anonymous · Answer

yep. :)

anonymous · Answer

Good (: so what is m<BAC?

anonymous · Answer

It readily follows, since ABC is an isosceles triangle. <BAC is the vertex angle, and <ABC and <BCA are base angles which are equal.