Question

In the given figure, O is the centre of the circle and LN is a diameter. If PQ is a tangent to the circle at K and angle KLN = 30 degree, find angle PKL.

Answer 1

Answer:

Step-by-step explanation:

Construction :- Draw OK (radius) to QP

IN Tri OKL,

OKL=OLK(Angle opposite to equal sides)

So, OKL =30

OKP =90(Tangent)

OLK+PKL=90

PKL=90-30=60

And that's all........Hope it helps you...

Answer 2

Angle PKL=30 degree

Step-by-step explanation:

In given figure

LN is a diameter of circle.

Angle KLN=30 degree

Construction :Join OK and OK is perpendicular to LN

Angle LOK=90 degree

In triangle OLK

$\angle OLK+\angle LKO+\angle LOK=180^{\circ}$

Using triangle angles sum property

Substitute the values

$30+90+\angle LKO=180$

$\angle LKO=180-30-90=60^{\circ}$

We know that

Radius is always perpendicular to tangent.

$\angle OKP=90^{\circ}$

$\angle PKL=\angle OKP-\angle LKO=90-60=30^{\circ}$

Hence, the angle PKL=30 degree

#Learns more:

https://brainly.in/question/7987689