Not the exact question you were looking for?
Paste your question to our Mathbee AI Mentor below to get an instant step-by-step solution.
Circles
CBSE 2026 Board Exam Set 1 (Code 30/7/1)
CBSE_BOARD_PYQ_2026_30_7_1
Grade 10
Question:
[Section C]
In the given figure, $XY$ and $X'Y'$ are two parallel tangents to a circle with center $O$ and another tangent $AB$ with point of contact $C$ intersecting $XY$ at $A$ and $X'Y'$ at $B$. Prove that $\angle AOB = 90^\circ$.
In the given figure, $XY$ and $X'Y'$ are two parallel tangents to a circle with center $O$ and another tangent $AB$ with point of contact $C$ intersecting $XY$ at $A$ and $X'Y'$ at $B$. Prove that $\angle AOB = 90^\circ$.
Step-by-Step Solution
Key Concept: Congruence of triangles $\triangle APO \cong \triangle ACO$ and $\triangle BQO \cong \triangle BCO$.
Join $OC$. $\triangle APO \cong \triangle ACO \Rightarrow \angle 1 = \angle 2$. [1.0 Mark]
$\triangle BQO \cong \triangle BCO \Rightarrow \angle 4 = \angle 3$. [1.0 Mark]
Since $POQ$ is a straight line, $2\angle 2 + 2\angle 3 = 180^\circ \Rightarrow \angle 2 + \angle 3 = 90^\circ \Rightarrow \angle AOB = 90^\circ$. Hence proved. [1.0 Mark]
$\triangle BQO \cong \triangle BCO \Rightarrow \angle 4 = \angle 3$. [1.0 Mark]
Since $POQ$ is a straight line, $2\angle 2 + 2\angle 3 = 180^\circ \Rightarrow \angle 2 + \angle 3 = 90^\circ \Rightarrow \angle AOB = 90^\circ$. Hence proved. [1.0 Mark]
Correct Answer: Proof of right angle subtended by tangent segment.
Mathbee AI Mentor (Free Demo)
Confused by the solution? Ask the AI to explain a specific step, tell you where you went wrong, or break down the key trap in this question.