Question

Given : PK and PL - tangent

Answer 1

Answer:

∠KPL = 80°

Step-by-step 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°

Answer 2

As PK=PL

• KM perpendicular to KP
• <PKM=90°

As

triangles are similar it's a inscribed right angle triangle .

Hence

• <KLM=90°

So

• <KML=50°

• <LKM be x

Use angle sum property

• x+50+90=180
• x+140=80
• x=40°

So

• <LKM+<PKL=90°
• 40+<PKL=90°
• <PKL=50°

And

• <PLK=50°(PK=PL hence isosceles triangle)

Hence

• <PKL=180-(50+50)
• <PKL=180-100
• <PKL=80°