Question 17 - CBSE Class 10 Sample Paper for 2027 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards

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Question 17 A flying kite is tied to a point on the ground with the help of a string. The string makes an angle 𝜃 with the ground level such that tan⁡𝜃=(12)/(5). If the length of the string is 52 m, then the height of the kite above the ground is: (A) 40 m (B) 45.5 m (C) 48 m (D) 50 m Let height of kite be AB And string be AC Making angle of 𝜃 with horizontal Now, AC = 52 m tan⁡𝜃=(12)/(5) We need to find AB We know that sin⁡𝜃=(Side opposite angle 𝜃 )/(Hypotenuse) 𝐬𝐢𝐧⁡𝜽=(𝑨𝑩)/(𝐀𝐂) Finding sin 𝜃 from tan 𝜃 We know that 𝟏+𝐭𝐚𝐧^(𝟐)𝜽=𝐬𝐞𝐜^(𝟐)𝜽 Putting 𝑠𝑒𝑐 𝜃=(1)/(𝑐𝑜𝑠 𝜃 ) 1+tan^(2)𝜃=(1)/(cos^(2)𝜃 ) Putting 𝑐𝑜𝑠^(2)𝜃=1−𝑠𝑖𝑛^(2)𝜃 1+tan^(2)𝜃=(1)/(1 − sin^(2)𝜃) Putting tan𝜃=(12)/(5) 1+((12)/(5))^(2)=(1)/(1 − sin^(2)𝜃) 1+(144)/(25)=(1)/(1 − sin^(2)𝜃) (25 + 144)/(25)=(1)/(1 − sin^(2)𝜃) (169)/(25)=(1)/(1 − sin^(2)𝜃) (169)/(25)=(1)/(1 − sin^(2)𝜃) 𝜃 B C 52 m ? A Seppe We need to find AB We know that sin⁡𝜃=(Side opposite angle 𝜃 )/(Hypotenuse) 𝐬𝐢𝐧⁡𝜽=(𝑨𝑩)/(𝐀𝐂) Finding sin 𝜃 from tan 𝜃 Since 𝑡𝑎𝑛𝜃=(12)/(5) 𝒔𝒊𝒏𝜽=(𝟏𝟐)/(sqrt(𝟏𝟐^(𝟐) + 𝟓^(𝟐)))=(12)/(sqrt(144 + 25))=(12)/(sqrt(169))=(𝟏𝟐)/(𝟏𝟑) 𝒄𝒐𝒔𝜽=(𝟓)/(sqrt(𝟏𝟐^(𝟐) + 𝟓^(𝟐)))=(12)/(sqrt(144 + 25))=(12)/(sqrt(169))=(𝟏𝟐)/(𝟏𝟑) 𝜃 B C 52 m ? A Seppe Or we can do the long method We know that 𝟏+𝐭𝐚𝐧^(𝟐)𝜽=𝐬𝐞𝐜^(𝟐)𝜽 Putting 𝑠𝑒𝑐 𝜃=(1)/(𝑐𝑜𝑠 𝜃 ) 1+tan^(2)𝜃=(1)/(cos^(2)𝜃 ) Putting 𝑐𝑜𝑠^(2)𝜃=1−𝑠𝑖𝑛^(2)𝜃 1+tan^(2)𝜃=(1)/(1 − sin^(2)𝜃) Putting tan𝜃=(12)/(5) 1+((12)/(5))^(2)=(1)/(1 − sin^(2)𝜃) 1+(144)/(25)=(1)/(1 − sin^(2)𝜃) 𝜃 B C 52 m ? A Seppe (25 + 144)/(25)=(1)/(1 − sin^(2)𝜃) (169)/(25)=(1)/(1 − sin^(2)𝜃) 1−sin^(2)𝜃=(25)/(169) 1−(25)/(169)=sin^(2)𝜃 (169 − 25)/(169)=sin^(2)𝜃 (144)/(169)=sin^(2)𝜃 sin^(2)𝜃=(144)/(169) 𝐬𝐢𝐧 𝜽=(𝟏𝟐)/(𝟏𝟑) 𝜃 B C 52 m ? A Seppe Now, 𝑠𝑖𝑛⁡𝜃=(𝐴𝐵)/(AC) Putting values (𝟏𝟐)/(𝟏𝟑)=(𝑨𝑩)/(𝟓𝟐) (12)/(13) × 52=𝐴𝐵 12 × 4 = AB 48 = AB AB = 48 m So, the correct answer is (C) 𝜃 B C 52 m ? A Seppe