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:
2b+(180−2a)=a⇒3a−2b=180
. . . we have make the top two equations equal to 180 and simplify.
.Subtract and angle b is 180/7
