The diagram below shows circle LMNP with KL a tangent to the circle at L - NSC Technical Mathematics - Question 8 - 2021 - Paper 2

The angle ∠P₂ is subtended by arc KL on the opposite side of the circle. According to the Inscribed Angle Theorem, the measure of an inscribed angle is half the measure of the arc it subtends.

Since ∠M = 98°,

$ext{∠P₂} = 180° - ∠M = 180° - 98° = 82°.$

Thus, ∠P₂ = 82°.

Since ∠P₁ and ∠P₂ are on the same straight line, we have:

$ext{∠P₁} + ext{∠P₂} = 180°.$

Substituting ∠P₂ = 82° into the equation gives:

$ext{∠P₁} + 82° = 180° \\ ext{∠P₁} = 180° - 82° = 98°.$

Therefore, ∠P₁ = 98°.

Using the property of tangents to circles, we know that the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment. Thus:

$ext{∠L₁} = ∠N₁ = 27°.$

Therefore, ∠L₁ = 27°.

From the previous calculations, we established that ∠KLP = ∠KNI, as they are corresponding angles due to the transversal line LN. Thus, by the Alternate Interior Angles Theorem, we conclude that Δ KLP || Δ KNl.

Using the property of segments in similar triangles derived from the previous proof of parallel lines:

This relation holds true as:

$KL² = KN·KP$

Substituting known values gives:

(6)² = 13 imes KP \\ 36 = 13·KP \\ KP = rac{36}{13} \\ KP ≈ 2.77 units.

For KLNM to be a cyclic quadrilateral, the opposite angles must sum to 180°. Using the calculated angles:

$K + M = 27° + 98° = 125°; \\ L + N = 55° + 82° = 137°;$

Since 125° ≠ 180° and 137° ≠ 180°, KLNM cannot be a cyclic quadrilateral.