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

To solve this problem, we need to use some properties of circles and angles.

Step 1: Recall that in a circle, the angle formed by a tangent and a chord is equal to the angle formed by the chord and the arc it intersects.

Step 2: In this case, we are given that LN is a diameter of the circle. This means that angle LKN is a right angle (90 degrees).

Step 3: We are also given that angle KLN is 30 degrees plus some unknown angle. Let's call this unknown angle x. So, angle KLN = 30 + x degrees.

Step 4: Since angle LKN is a right angle, angle LKN + angle KLN = 90 degrees. Substituting the values we know, we have 30 + x + 90 = 90 degrees.

Step 5: Simplifying the equation, we have 30 + x = 0 degrees.

Step 6: Solving for x, we subtract 30 from both sides of the equation. This gives us x = -30 degrees.

Step 7: Now that we know the value of x, we can find the angle PKL. Since PK is a tangent to the circle, angle PKL is equal to angle KLN. So, angle PKL = 30 + x = 30 - 30 = 0 degrees.

Step 8: Therefore, the value of angle PKL is 0 degrees.