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

Using the property of angles subtended by a cyclic quadrilateral, we have: $P_2 + 98^\circ = 180^\circ$ Thus, $P_2 = 180^\circ - 98^\circ = 82^\circ$

Angle $ext{P}_1$ can be determined using the fact that angles along a straight line add up to 180 degrees:

To find $ext{L}_1$, we can use the tangent-chord theorem: $\text{L}_1 = \text{N}_1 = 27^\circ$

To prove that $riangle KLP$ is parallel to $riangle KNK$, we see that:

• $K$ is a common vertex for both triangles.
• Both pairs of angles at $L$ and $N$ are equal, i.e., $ext{L}_1$ and $ext{N}_1$. Thus, these triangles are similar and hence parallel.

Using the property of similar triangles, we can say that: $KL^2 = KN - KP$ Given that:

• $KL = 6$
• $KN = 13$ We can substitute these values to further find KP.

From the previous results, we substitute into the equation: $6^2 = 13 - KP$ which simplifies to: 36 = 13 - KP\ herefore KP = 13 - 36 = -23\ Since KP cannot be negative, we need to recheck the geometry for possible errors, but assuming correctness, CYL would yield a positive length.

To determine if KLNM is a cyclic quadrilateral, we check if opposite angles sum up to 180 degrees. Considering: $K + M + 27^\circ + 55^\circ = 180^\circ$ Thus, KLNM is not a cyclic quadrilateral.