If a angle between two tangents drawn from a point $P$ to a circle of radius $r$ and center $O$ is $2\theta$, prove that:
(i) $TP = r \cot \theta$
(ii) $OP = r \csc \theta$
(iii) Area of quadrilateral $OPTQ = r^2 \cot \theta$.
Step-by-Step Solution
Key Concept: In right $\Delta OPT$: angle at $P$ is $\theta$, angle at $T$ is $90^\circ$, angle at $O$ is $90^\circ - \theta$.
Stepwise Solution:
Since $OP$ bisects the angle between tangents $\angle TPT' = 2\theta$, in right $\Delta OPT$, $\angle OPT = \theta$ and $\angle OTP = 90^\circ$. [1.0 Mark]
(i) $\cot \theta = \dfrac{TP}{OT} = \dfrac{TP}{r} \Rightarrow TP = r \cot \theta$. Proved! [1.5 Marks]
(ii) $\csc \theta = \dfrac{OP}{OT} = \dfrac{OP}{r} \Rightarrow OP = r \csc \theta$. Proved! [1.5 Marks]
(iii) $\text{Area}(OPTQ) = 2 \times \text{Area}(\Delta OPT) = 2 \times \left(\dfrac{1}{2} \times r \times r \cot \theta\right) = r^2 \cot \theta$. Proved! [1.0 Mark]
Marking Scheme:
• Setting angle $\angle OPT = \theta$ in right $\Delta OPT$: 1.0 Mark
• Proving Part (i) $TP = r \cot \theta$: 1.5 Marks
• Proving Part (ii) $OP = r \csc \theta$: 1.5 Marks
• Proving Part (iii) $\text{Area}(OPTQ) = r^2 \cot \theta$: 1.0 Mark
Correct Answer: