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Transcript

Ex 8.1, 9 In triangle ABC, right-angled at B, if tan A = 1/√3, find the value of sin A cos C + cos A sin C Given tan A = 1/√3 (π‘Ίπ’Šπ’…π’† π’π’‘π’‘π’π’”π’Šπ’•π’† 𝒕𝒐 𝑨)/(π‘Ίπ’Šπ’…π’† 𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕 𝒕𝒐 𝑨) = 𝟏/βˆšπŸ‘ 𝐡𝐢/𝐴𝐡 = 1/√3 Let BC = x & AB = βˆšπŸ‘ x We have to find sin A cos C + cos A sin C Putting sin A = 1/2 , cos A = √3/2 , sin C = √3/2 & cos C = 1/2 = (𝟏/𝟐)Γ—(𝟏/𝟐)+(βˆšπŸ‘/𝟐)Γ—(βˆšπŸ‘/𝟐) = 1/4 + (√3 Γ— √3)/4 = 1/4 + 3/4 = (1 + 3)/4 = 4/4 = 1 So, sin A cos C + cos A sin C = 1 Ex 8.1, 9 In triangle ABC, right-angled at B, if tan A = 1/√3, find the value of (ii) cos A cos C – sin A sin C cos A cos C – sin A sin C Putting sin A = 1/2 , cos A = √3/2 , sin C = √3/2 & cos C = 1/2 = (βˆšπŸ‘/𝟐)Γ—πŸ/πŸβˆ’(𝟏/𝟐)Γ—(βˆšπŸ‘/𝟐) = (√3/4)βˆ’(√3/4) = 0 Hence, cos A cos C – sin A sin C = 0

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo