Answer:
∠KPL = 80°
Explanation:
As KM is perpendicular to KP, this means that KM is the diameter of the circle, as the tangent of a circle is always perpendicular to the radius.
The angle at the circumference in a semicircle is always a right angle. So if KM is the diameter then ΔKLM is a right triangle with ∠KLM = 90°
The sum of the interior angles of a triangle is 180°.
Therefore, ∠KLM + ∠KML + ∠LKM = 180°
As ∠KLM = 90° and ∠KML = 50°
⇒ 90° + 50° + ∠LKM = 180°
⇒ ∠LKM = 40°
As ∠PKM = 90° then
⇒ ∠PKL = 90 - ∠LKM
= 90 - 40
= 50°
Tangents from a common point to a circle are always equal in length, so as PK and PL are tangents from point P, then PK = PL
Therefore, ΔLPK is an isoceles triangle and so its base angles ∠PKL and ∠PLK will be equal.
Since ∠PKL = 50° then ∠PLK = 50°
Again, as the sum of interior angles of a triangle is 180°:
⇒ ∠KPL + ∠PKL + ∠PLK = 180°
⇒ ∠KPL + 50° + 50° = 180°
⇒ ∠KPL = 80°