All give you a good starting point.
Let angle b = x
Now you can get every angle there in terms of x.
Start with the most obvious and work from there.
It is not all that hard.
eventually you will have 2 or 3 angles all in terms of x that add to give 180 degrees.
Then you will just have a little equation to solve to get x.
When you finish you should check that it works,
Please no one else give a full answer.
Jenny, you can interact with me if you need/want to.
Should i label other angles a different variable (ig. angle p is y angle q is z)
No, you do not need to.
There are lots of isosceles triangles in there.
You can label all the angles in terms of x.
You can use
1. base angles in isosceles triangles,
2. angle sum of triangle
3. supplementary angles
I thank that is all you really need to use.
i did something like this, correct me if I am wrong. ;-;
In isosceles triangle APC, let a=∠ACP=∠APCAPC, let a=∠ACP=∠APC
. . Then:.∠PAC=180−2a∠PAC=180−2a
In isosceles triangle QBP, let b=∠QBP=∠QPBQBP, let b=∠QBP=∠QPB
Since ∠AQP∠AQP is an exterior angle: ∠AQP=2b∠AQP=2b
In isosceles triangle APQ,∠QAP=2b⇒∠QPA=180−4bAPQ,∠QAP=2b⇒∠QPA=180−4b
At point P:∠BPQ+∠QPA+∠APC=180P:∠BPQ+∠QPA+∠APC=180
. . Hence: b +(180-4b) +a=180 ⇒a-3b=0
At point A:∠QAP+∠PAC=∠QAC=∠ACPA:∠QAP+∠PAC=∠QAC=∠ACP
. . Hence:
. . . we have make the top two equations equal to 180 and simplify.
.Subtract and angle b is 180/7
I'm not going to check your whole answer, and I have thrown out my working now, but that does sound familiar.