Question 37 - CBSE Class 10 Sample Paper for 2027 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards

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Question 37 In a technology park in Hyderabad, a company installed a circular meditation garden for its employees as shown in the figure. Arjun, a designer, stood at a point P outside the garden and drew two tangents PA and PB to the circular boundary. The radius of the garden is 9 m, and the distance of point P from the centre O is 18 m. For preparing safety guidelines, Arjun needed the lengths of the tangents. During a demonstration to interns, he asked them to compute the angle formed between the two tangents using the properties of circle. Based on the above information, answer the following questions: (i) What is the measure of ∠OAP? Based on the above information, answer the following questions: (i) What is the measure of ∠OAP? Here, PA, PB is tangent to circle and OA, OB is radius We know that Tangent is perpendicular to radius ∴ OA⊥ AP Thus, ∠ OAP = 90° Question 37 (ii) (ii) Calculate the length of the tangent PA. Given Radius = OA = OB = 9 m Distance from O to P = OP = 18 m In right triangle ∆ OAP By Pythagoras Theorem 𝑂𝑃^(2)=𝐴𝑃^(2)+𝑂𝐴^(2) 9 m 18 m 𝟏𝟖^(𝟐)=𝐀𝐏^(𝟐)+𝟗^(𝟐) 324=AP^(2)+81 324−81=AP^(2) 324=AP^(2)+81 243=AP^(2) 𝑨𝑷^(𝟐)=𝟐𝟒𝟑 𝐴𝑃=sqrt(243) 𝐴𝑃=sqrt(81 × 3) 𝐴𝑃=sqrt(9^(2) × 3) 𝑨𝑷=𝟗sqrt(𝟑) m Question 37 (iii) – (A) Calculate the measure of ∠AOB. By symmetry ∠ AOP = ∠ BOP = (𝟏)/(𝟐)∠ AOB Let ∠ AOP = 𝜃 We need to find ∠ AOB = 2𝜃 Now, 9 m 18 m 9sqrt(3) m 𝜃 sin 𝜃 = (𝑺𝒊𝒅𝒆 𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆 𝒕𝒐 𝒂𝒏𝒈𝒍𝒆 𝜽)/(𝑯𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒆) sin 𝜃 = (𝐴𝑃)/(𝑂𝑃) sin 𝜃 = (9sqrt(3))/(18) sin 𝜃 = (sqrt(𝟑))/(𝟐) Since sin 60° = (sqrt(3))/(2) ∴ 𝜃 = 60° So, ∠ AOB = 2𝜃 = 2 × 60° = 120° Question 37 (iii) – (B) (iii) In the above design, if PA and PB were inclined at 60°, then what would be their length? Given ∠ APB = 60° By symmetry ∠ APO = (𝟏)/(𝟐)∠ APB = (1)/(2) × 60° = 30° 9 m 60° 18 m 30° Now, in ∆ APO tan P = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑎𝑛𝑔𝑙𝑒 𝑃)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝑃) tan 30° = (𝑶𝑨)/(𝑷𝑨) (1)/(sqrt(3)) = (9)/(𝑃𝐴) PA = 9sqrt(𝟑) m