Answer:
55°
Step-by-step explanation:
In circle with center O, OA and OB are radii and PA and PB are tangents drawn from external point P.
Since, tangent is perpendicular to the radius of the circle.
[tex] \therefore m\angle PAO = m\angle PBO = 90°\\
m\angle APB = 70°....(GIVEN) \\[/tex]
In quadrilateral PAOB,
[tex] m\angle PAO + m\angle PBO+ m\angle APB \\+m\angle AOB = 360°\\
\therefore 90° + 90° + 70° +m\angle AOB = 360°\\
\therefore 250° +m\angle AOB = 360°\\
\therefore m\angle AOB = 360°- 250°\\
\huge \purple {\boxed {\therefore m\angle AOB = 110°}} \\[/tex]
Since, angle subtended at the circumference of the circle is half of the angle subtended at the centre of the circle.
[tex] \therefore m\angle ACB = \frac {1}{2} \times m\angle AOB\\\\
\therefore m\angle ACB = \frac {1}{2} \times 110°\\\\
\huge \orange {\boxed {\therefore m\angle ACB = 55°}} [/tex]
