Question

anyone please solve this problem

Answer 1

<ABC=180°-130°=50° (Linear Pair)
<CAB=<ABC. (CA=CB)
Therefore,2<CAB=180°-50°=130°
That is,<ABC=<CAB=130°/2=65°

<PAC=90°. (Theorem 10.2)
<PAC=<CAB+<BAP
90° =65°+<BAP

Therefore,<BAP=25°.

Hope it helps.
Please mark it as brainliest.

Answer 2

hello dear,

WE KNOW THAT AC=BC=RADIUS OF CIRCLE


HENCE, <ABC= <CAB



now we,.
know that, PBC is a straight line


then, <PBC=180


then,<ABP+<ABC=180


=>. <ABC=180-<ABP
=>. <ABC=180-130
=>. <ABC=50 =<CAB



NOW,. AP IS A TENGENT
HENCE, <PAB=90

IN,∆ABC,
<ACB+<CAB+<ABC=180. { <ABC=<CAB=50}

=>. <ACB=180-100

=>. <ACB=80

NOW, IN ∆APB


<ACP +<APC +<PAC=180


=>. 80 + 90+ <APC =180

=>. <APC =180-170

<APC=10


IN ∆APB,
<ABP + <APB + <PAB =180

=>. <PAB =180 - 10-130

=> <PAB=40 =<BAP


SO THE ANSWER IS <BAP =40


I HOPE ITS HELP YOU DEAR,
THANKS