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In Fig. 2, O is the centre of the circle and LN is a diameter. If PQ is a tangent to the circle at K and $\angle K\!LN=30^{\circ}$ , find $\angle P\!K\!L$ .

Answers (1)

Given : $\angle K\!LN=30^{\circ}$

In $\bigtriangleup LK\!N$ ,

$\angle LK\!N = 90^{\circ}$ ( angle subtended by diameter on the circumference of the circle is $90^{\circ}$ )

Also, $OK=$ Radius of the circle.

$\angle O\!K\!P=90^{\circ}\;\;\;\;\; -(1)$ (tangent at any point on the circle is perpendicular to the radius through the point of contact)

In $\bigtriangleup LO\!K$ ,

$\angle LO\!K=90^{\circ}$

$\angle K\!LN=30^{\circ}$ (Given)

$\angle LK\!O + \angle LO\!K + \angle K\!L\!N=180^{\circ}$ (Sum of angles of a triangle)

$\angle LK\!O + 90^{\circ} + 30^{\circ}=180^{\circ}$

$\angle LK\!O = 180^{\circ} + 120^{\circ}=60^{\circ}$

$\angle O\!K\!P = 90^{\circ}$ - (from (1))

$\angle O\!K\!P = \angle LK\!O + \angle P\!K\!L = 90^{\circ}$

$\Rightarrow 60^{\circ} + \angle P\!K\!L = 90^{\circ}$

$\Rightarrow \angle P\!K\!L = 90^{\circ} - 60^{\circ}$

$\Rightarrow \angle P\!K\!L = 30^{\circ}$

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